1001

using namespace std中存在std::count，所以再定义count会报错：count不明确

#include <iostream>
using namespace std;

int n, cnt = 0;

void cal(int num) {
    if (num == 1)
    {
        return;
    }
    if (num % 2 == 0)
    {
        cnt++;
        cal(num / 2);
        return;
    }
    cnt++;
    cal((3 * num + 1) / 2);
}

int main()
{
    cin >> n;
    cal(n);
    cout << cnt << endl;
    return 0;
}

1002

char转int方法：
可以定义一个宏#define charToNum(x) (x-'0')

#include <iostream>
#include <string>
#include <vector>
#define charToNum(x) (x-'0')
using namespace std;

string n;
vector<string> store;

void intToPY(int x) {
    switch (x) {
    case 1:
        store.push_back("yi");
        break;
    case 2:
        store.push_back("er");
        break;
    case 3:
        store.push_back("san");
        break;
    case 4:
        store.push_back("si");
        break;
    case 5:
        store.push_back("wu");
        break;
    case 6:
        store.push_back("liu");
        break;
    case 7:
        store.push_back("qi");
        break;
    case 8:
        store.push_back("ba");
        break;
    case 9:
        store.push_back("jiu");
        break;
    case 0:
        store.push_back("ling");
        break;
    }
}

void print(int sum) {
    if (sum == 0)
        return;
    int temp = sum % 10;
    print(sum/10);
    intToPY(temp);
}

void solve() {
    if (n == "0")
    {
        cout << "ling" << endl;
        return;
    }
    int sum = 0;
    for (int i = 0; i < n.size(); ++i) {
        sum += charToNum(n[i]);
    }
    print(sum);
    for (int i = 0; i < store.size() - 1; ++i) {
        cout << store[i] << " ";
    }
    cout << store[store.size() - 1] << endl;
}

int main()
{
    cin >> n;
    solve();
    return 0;
}

然而我的算法过于复杂，其实可以这么写：
输入无需储存起来，可以使用(c=getchar())!=\n一个char一个char的扫描
由于0~9与数组下标0~9一一对应，可以定义一个字符串数组用于储存0~9的拼音。
由于最大和为9*100<1000，所以只需取余三次即可。

#include <stdio.h>
#include <string.h>
#define charToNum(x) (x-'0')

int main()
{
    char c;
    int sum = 0;
    while ((c = getchar()) != '\n') sum += charToNum(c);
    const char *intToPY[] = { "ling","yi","er","san","si","wu","liu","qi","ba","jiu" };
    if (sum / 100)
        printf("%s ", intToPY[sum / 100]);
    if (sum / 10)
        printf("%s ", intToPY[sum / 10 % 10]);
    printf("%s\n", intToPY[sum % 10]);
    return 0;
}

1003

给定一个字符串，要求能化简成xPATx形式（x为只含A的字符串或者为空字符串）
如果 aPbTc 是正确的，那么 aPbATca 也是正确的，其中 a、 b、 c 均或者是空字符串，或者是仅由字母 A 组成的字符串。
设P前A的个数为x，P与T中间A的个数为y，T之后A的个数为z，
所以只需满足递推式，并且

#include <iostream>
#include <string>
#include <vector>
using namespace std;

int n;
void solve(string s) {
    int x = 0, y = 0, z = 0;
    int pCount = 0, tCount = 0;
    for (int i = 0; i < s.size(); ++i)
    {
        if ((s[i] != 'P'&&s[i] != 'A'&&s[i] != 'T')|| (s[i] == 'P' && (pCount>0 || tCount>0))||(s[i]=='T'&&tCount>0))
        {
            cout << "NO" << endl;
            return;
        }
        if (s[i] == 'P') pCount++;
        if (s[i] == 'T') tCount++;
    }
    if (pCount != 1 || tCount != 1) {
        cout << "NO" << endl;
        return;
    }
    int k = 0;
    while (s[k] != 'P'&&k<s.size()-1) {
        x++;
        k++;
    }k++;
    while (s[k] != 'T'&&k < s.size()) {
        y++;
        k++;
    }k++;
    while (k < s.size()) {
        z++;
        k++;
    }
    if (z - (y - 1)*x == x && y != 0)
    {
        cout << "YES" << endl;
        return;
    }
    else {
        cout << "NO" << endl;
        return;
    }
}

int main()
{
    cin >> n;
    vector<string> input(n);
    for (int i = 0; i < n; ++i) {
        cin >> input[i];
    }
    for (int i = 0; i < n; ++i) {
        solve(input[i]);
    }
    return 0;
}

1004

#include <iostream>
#include <string>
using namespace std;

int main()
{
    int n;
    int low = 100, high = 0;
    int point;
    string name, number;
    string highP, lowP;
    cin >> n;
    for (int i = 0; i < n; ++i) {
        cin >> name >> number >> point;
        if (point >= high) {
            highP = name + ' ' + number;
            high = point;
        }
        if (point <= low) {
            lowP = name + ' ' + number;
            low = point;
        }
    }
    cout << highP << endl << lowP << endl;
    return 0;
}

1005

第一题的复杂版
思路是每一次运算都与输入的整数比较一次，如果有相同的，说明它是这条链上的一个节点，
将其设为1表示删除

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
vector<int> input;

bool compare(int a, int b) {
    return a > b;
}

void del(int x) {
    if (x == 1)
        return;
    if (x % 2) {
        int k = (3 * x + 1) / 2;
        for (int i = 0; i < input.size(); ++i) {
            if (k == input[i])
            {
                input[i] = 1;
                break;
            }
        }
        del(k);
    }
    else {
        int k = x / 2;
        for (int i = 0; i < input.size(); ++i) {
            if (k == input[i])
            {
                input[i] = 1;
                break;
            }
        }
        del(k);
    }
}

void solve() {
    for (int i = 0; i < input.size(); ++i) {
        del(input[i]);
    }
    sort(input.begin(), input.end(), compare);
    int i;
    for (i= 0; i < input.size() && input[i] != 1; ++i) {}
    for (int k = 0; k < i - 1; k++) {
        cout << input[k] << " ";
    }
    cout << input[i - 1] << endl;
}

int main()
{
    int n;
    cin >> n;
    input.resize(n);
    int temp;
    for (int i = 0; i < n; ++i) {
        cin >> input[i];
    }
    solve();
    return 0;
}

1006

#include <iostream>
using namespace std;
int num;

void solve() {
    if (num / 100) {
        for (int i = 0; i < (num / 100); ++i) {
            cout << 'B';
        }
    }
    if (num % 100 / 10) {
        for (int i = 0; i < (num % 100 / 10); ++i) {
            cout << 'S';
        }
    }
    for (int i = 1; i <= (num % 10); ++i) {
        cout << i;
    }
    cout << endl;
}

int main()
{
    cin >> num;
    solve();
    return 0;
}

1007

memset的头文件为<string.h>，记得引入
思路是首先用埃氏筛法得到n以内所有的素数，然后用相邻项相减，记录差为2的素数对的个数。

#include <iostream>
#include <vector>
#include <string.h>
#define MAX_N 100010
using namespace std;
int n;
bool Primes[MAX_N];
vector<int> save;
int cnt = 0;

bool isPrime(int x) {
    for (int i = 2; i*i <= x; ++i) {
        if (x%i == 0)
            return false;
    }
    return true;
}

void solve() {
    memset(Primes, 0, sizeof(Primes));
    Primes[1] = Primes[0] = true;
    for (int i = 2; i <= n; ++i) {
        if (isPrime(i)) {
            for (int k = i; k <= n; k += i) {
                Primes[i] = true;
            }
        }
    }
    for (int i = 2; i <= n; ++i) {
        if (Primes[i])
            save.push_back(i);
    }
    for (int i = 0; i < save.size()-1; ++i) {
        if (save[i + 1] - save[i] == 2)++cnt;
    }
    cout << cnt << endl;
}

int main()
{
    cin >> n;
    solve();
    return 0;
}

1008

本题的关键是没有规定m一定小于n，所以要记得将m取n的模

#include <iostream>
#define MAX_N 110
using namespace std;
int n, m;
int Arr[MAX_N];

void swap(int &a, int &b) {
    int temp = a;
    a = b;
    b = temp;
}

void solve() {
    for (int i = n - m; i < n; ++i) {
        for (int j = i; j > i - n + m; --j) {
            swap(Arr[j], Arr[j - 1]);
        }
    }
    for (int i = 0; i < n - 1; i++)
        cout << Arr[i] << " ";
    cout << Arr[n-1] << endl;
}

int main()
{
    cin >> n >> m;
    m %= n;
    for (int i = 0; i < n; ++i) {
        cin >> Arr[i];
    }
    solve();
    return 0;
}

1009

本题的关键是要将每个单词保存下来，可以用c=getchar()检测每次的输入，
当输入字母时，将其与上一次输入相加。当输入空格时，先将字符串保存，再将字符串清空，并停止此次循环。
由于结尾没有空格，所以最后一个单词单独保存。

#include <iostream>
#include <vector>
#include <string>
using namespace std;

int main()
{
    char c;
    vector<string> save;
    string temp="";
    while ((c = getchar()) != '\n') {
        if (c == ' ')
        {
            save.push_back(temp);
            temp = "";
            continue;
        }
        temp += c;
    }
    save.push_back(temp);
    for (int i = save.size() - 1; i > 0; --i) {
        cout << save[i] << " ";
    }
    cout << save[0] << endl;
    return 0;
}

1010

本题的难点在于如何判断输入结束
while (scanf("%d %d", &coef, &index) != EOF)可以用于扫描一行的内容，EOF代表输入的末尾。

#include <stdio.h>

int main()
{
    int coef, index, cnt = 0, flag = 1;
    while (scanf("%d %d", &coef, &index) != EOF) {
        if (index > 0) {
            if (flag) {
                printf("%d %d", coef*index, index - 1);
                cnt++;
                flag = 0;
            }
            else {
                printf(" %d %d", coef*index, index - 1);
                cnt++;
            }
        }
    }
    if (cnt == 0) {
        printf("0 0");
    }
    return 0;
}

1011

由于输入A,B,C的范围为~，int占2个字节16位，long占4个字节32位，所以不能用int，要使用long。

#include <iostream>
#define MAX_N 20
using namespace std;
int n;
bool save[MAX_N];

int main()
{
    cin >> n;
    long a, b, c;
    for (int i = 1; i <= n; ++i) {
        cin >> a >> b >> c;
        a + b > c ? save[i] = true : save[i] = false;
    }
    for (int i = 1; i <= n; ++i) {
        cout << "Case #" << i << ": ";
        save[i] ? cout << "true" : cout << "false";
        cout << endl;
    }
    return 0;
}

1012

对于printf("%x,yf",a)，若a的整数部分不足x位，则在左边补零，若超过x位，则正常输出，若a的小数部分不足y位，则在右边补零，若超过y位，则四舍五入保留y位小数。

#include <iostream>
#include <string.h>
#include <algorithm>
#include <stdio.h>
#define MAX_N 1010
using namespace std;
int N;
int save[MAX_N];
int output[5];

void solve() {
    bool flag = true;
    int count = 0;
    bool visit[5] = { false };
    memset(output, 0, sizeof(output));
    for (int i = 0; i < N; ++i) {
        if (save[i] % 5 == 0 && save[i] % 2 == 0) {
            output[0] += save[i];
            visit[0] = true;
        }
        else if (save[i] % 5 == 1) {
            flag ? output[1] += save[i] : output[1] -= save[i];
            flag = !flag;
            visit[1] = true;
        }
        else if (save[i] % 5 == 2) {
            output[2] += 1;
            visit[2] = true;
        }
        else if (save[i] % 5 == 3) {
            output[3] += save[i];
            ++count;
            visit[3] = true;
        }
        else if (save[i] % 5 == 4) {
            output[4] = max(output[4], save[i]);
            visit[4] = true;
        }
        else {
            visit[0] == true;
        }
    }
    for (int i = 0; i < 4; ++i) {
        if (visit[i] == false) {
            cout << "N" << " ";
        }
        else if (i != 3) {
            cout << output[i] << " ";
        }
        else if (i == 3) {
            printf("%.1f ", (float)output[i] / count);
        }
    }
    visit[4] ? cout << output[4] << endl : cout << "N" << endl;
}

int main()
{
    cin >> N;
    for (int i = 0; i < N; ++i) {
        cin >> save[i];
    }
    solve();
    return 0;
}

1013

2是最小的素数，1不是素数

#include <iostream>
#define MAX_S 10000010
using namespace std;
int lb, rb;
int save[MAX_S];

bool is_prime(int x) {
    for (int i = 2; i*i <= x; ++i) {
        if (x%i == 0)
            return false;
    }
    return true;
}

void solve() {
    int count = 1;
    int index = 3;
    save[1] = 2;
    while (count < rb) {
        if (is_prime(index)) {
            save[++count] = index;
        }
        index++;
    }
    while (rb - lb >= 9) {
        int i;
        for (i = lb; i < lb + 9; ++i) {
            cout << save[i] << " ";
        }
        cout << save[i] << endl;
        lb += 10;
    }
    if (lb <= rb) {
        for (int i = lb; i < rb; ++i) {
            cout << save[i] << " ";
        }
        cout << save[rb] << endl;
    }
}

int main()
{
    cin >> lb >> rb;
    solve();
    return 0;
}

1014

本题的难点在于需要熟悉ASCII

#include <iostream>
#include <algorithm>
#include <string>
#include<stdio.h>
#include <vector>
#define charToNum(x) (x-'A')
using namespace std;
vector<string> input(4);

int main()
{
    int flag = 0;
    const char *week[] = { "MON","TUE","WED","THU","FRI","SAT","SUN" };
    for (int i = 0; i < 4; ++i) {
        cin >> input[i];
    }
    int index1= min(input[0].size(), input[1].size());
    for (int i = 0; i < index1; ++i) {
        if (input[0][i] == input[1][i]) {
            if (flag == 0 && input[0][i] >= 'A'&&input[0][i] <= 'G') {
                cout << week[charToNum(input[0][i])] << " ";
                flag++;
            }
            else if (flag == 1 && input[0][i] == input[1][i]) {
                if (input[0][i] >= '0' && input[0][i] <= '9') {
                    printf("%02d:", input[0][i]-'0');
                    flag++;
                }
                else if (input[0][i] >= 'A' && input[0][i] <= 'N') {
                    cout << charToNum(input[0][i])+10 << ":";
                    flag++;
                }
            }
        }
    }
    int index2 = min(input[2].size(), input[3].size());
    for (int i = 0; i < index2; ++i) {
        if (input[2][i] == input[3][i]&&((input[2][i]>='a'&&input[2][i]<='z')|| (input[2][i] >= 'A'&&input[2][i] <= 'Z'))) {
            printf("%02d\n", i);
            break;
        }
    }
    return 0;
}

1015

此题思路还是很简单的，但是由于数据量较大，用cin、cout会超时。

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int N, L, H;
struct stu {
    int id;
    int de;
    int cai;
};

bool compare(stu a, stu b) {
    if (a.de + a.cai == b.de + b.cai) {
        if (a.de == b.de)
            return a.id < b.id;
        return a.de > b.de;
    }
    return a.de + a.cai > b.de + b.cai;
}

int main()
{
    int count = 0;
    scanf("%d %d %d", &N, &L, &H);
    vector<stu> people1;
    vector<stu> people2;
    vector<stu> people3;
    vector<stu> people4;
    stu pig;
    for (int i = 0; i < N; ++i) {
        scanf("%d %d %d", &pig.id, &pig.de, &pig.cai);
        if (pig.de >= L && pig.cai >= L) {
            if (pig.de >= H && pig.cai >= H) {
                people1.push_back(pig);
            }
            else if (pig.de >= H && pig.cai < H) {
                people2.push_back(pig);
            }
            else if (pig.de < H&&pig.cai < H&&pig.de>=pig.cai) {
                people3.push_back(pig);
            }
            else {
                people4.push_back(pig);
            }
            count++;
        }
    }
    sort(people1.begin(), people1.end(), compare);
    sort(people2.begin(), people2.end(), compare);
    sort(people3.begin(), people3.end(), compare);
    sort(people4.begin(), people4.end(), compare);
    cout << count << endl;
    for (int i = 0; i < people1.size(); ++i) {
        printf("%d %d %d\n", people1[i].id, people1[i].de, people1[i].cai);
    }
    for (int i = 0; i < people2.size(); ++i) {
        printf("%d %d %d\n", people2[i].id, people2[i].de, people2[i].cai);
    }
    for (int i = 0; i < people3.size(); ++i) {
        printf("%d %d %d\n", people3[i].id, people3[i].de, people3[i].cai);
    }
    for (int i = 0; i < people4.size(); ++i) {
        printf("%d %d %d\n", people4[i].id, people4[i].de, people4[i].cai);
    }
    return 0;
}

1016

计算次方的函数为pow()

#include <iostream>
#include <math.h>
#include<string>
#define charToNum(x) (x-'0')
using namespace std;

string A, B;
char Da, Db;

int main()
{
    cin >> A >> Da >> B >> Db;
    int num1 = 0, num2 = 0;
    for (int i = 0; i < A.size(); ++i) {
        if (A[i] == Da)++num1;
    }
    for (int i = 0; i < B.size(); ++i) {
        if (B[i] == Db)++num2;
    }
    int sum = 0;
    for (int i = 0; i < num1; ++i) {
        sum += (pow(10, i)*charToNum(Da));
    }
    for (int i = 0; i < num2; ++i) {
        sum += (pow(10, i)*charToNum(Db));
    }
    cout << sum << endl;
    return 0;
}

1017

由于被除数最高一千位，所以肯定需要使用高精度计算。

#include <iostream>
#include <string.h>
#include <string>
#define charToNum(x) (x-'0')
using namespace std;


int main()
{
    string a;
    int b, re = 0;
    int save[1010];
    int result[1010];
    memset(save, 0, sizeof(save));
    memset(result, 0, sizeof(result));
    cin >> a >> b;
    save[0] = a.size();
    for (int i = 0; i < a.size(); ++i) {
        save[i] = charToNum(a[i]);
    }
    for (int i = 0; i < a.size(); ++i) {
        result[i] = (re * 10 + save[i]) / b;
        re = (re * 10 + save[i]) % b;
    }
    int lenc = 0;
    while (result[lenc] == 0 && lenc < a.size()-1)lenc++;
    for (int i = lenc; i < a.size(); ++i) {
        cout << result[i];
    }
    cout <<" "<< re << endl;
    return 0;
}

1018

#include <iostream>
using namespace std;
int N;
struct pig {
    int scissors = 0;
    int stone = 0;
    int cloth = 0;
};

int sum(pig a) {
    return a.cloth + a.scissors + a.stone;
}

char big(pig a) {
    return a.scissors > a.stone ? (a.scissors > a.cloth ? 'J' : 'B') : (a.stone > a.cloth ? 'C' : 'B');
}

int main()
{
    pig p1, p2;
    int ping = 0;
    char temp1, temp2;
    cin >> N;
    for (int i = 0; i < N; ++i) {
        cin >> temp1 >> temp2;
        if (temp1 == temp2)ping++;
        else if (temp1 == 'C'&&temp2 == 'J')p1.stone++;
        else if (temp1 == 'J'&&temp2 == 'C')p2.stone++;
        else if (temp1 == 'B'&&temp2 == 'C')p1.cloth++;
        else if (temp1 == 'C'&&temp2 == 'B')p2.cloth++;
        else if (temp1 == 'J'&&temp2 == 'B')p1.scissors++;
        else if (temp1 == 'B'&&temp2 == 'J')p2.scissors++;
    }
    cout << sum(p1) << " " << ping << " " << sum(p2) << endl;
    cout << sum(p2) << " " << ping << " " << sum(p1) << endl;
    cout << big(p1) << " " << big(p2) << endl;
    return 0;
}

1019

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <math.h>
#include <string>
#define charToNum(x) (x-'0')
#define numToChar(x) (x+'0')
using namespace std;

int main()
{
    string a;
    cin >> a;
    for (int i = a.size(); i < 4; i++) {
        a += '0';
    }
    if (a[0] == a[1] && a[1] == a[2] && a[2] == a[3]) {
        cout << a << " - " << a << " = " << "0000" << endl;
    }
    else {
        do{
            sort(a.begin(), a.end());
            int a1 = 0, b = 0;
            for (int i = 0; i <= 3; ++i) {
                a1 += charToNum(a[i])*pow(10, i);
            }
            for (int i = 0; i <= 3; ++i) {
                b += charToNum(a[i])*pow(10, 3 - i);
            }
            int c = a1 - b;
            printf("%04d - %04d = %04d\n", a1, b, c);
            a = "";
            a += numToChar(c / 1000);
            a += numToChar(c % 1000 / 100);
            a += numToChar(c % 100 / 10);
            a += numToChar(c % 10);
        }while (a != "6174");
    }
    return 0;
}

1020

要看清题目描述哦....重量和价格不一定是整数

#include <iostream>
#include <stdio.h>
#include <algorithm>
using namespace std;
float N, D;
struct pie {
    float weight;
    float money;
    float ave;
};

bool compare(pie a, pie b) {
    return a.ave > b.ave;
}

int main()
{
    float sum = 0;
    cin >> N >> D;
    vector<pie> save(N);
    for (int i = 0; i < N; ++i) {
        cin >> save[i].weight;
    }
    for (int i = 0; i < N; ++i) {
        cin >> save[i].money;
    }
    for (int i = 0; i < N; ++i) {
        save[i].ave = save[i].money / save[i].weight;
    }
    sort(save.begin(), save.end(), compare);
    for (int i = 0; i < N; ++i) {
        if (D >= save[i].weight) {
            D -= save[i].weight;
            sum += save[i].money;
        }
        else {
            sum += save[i].ave*D;
            break;
        }
    }
    printf("%.2f\n", sum);
    return 0;
}

1021

#include <string>
#include <string.h>
#include <iostream>
#define charToNum(x) (x-'0')
using namespace std;
int a[10];

int main()
{
    memset(a, 0, sizeof(a));
    char c;
    while ((c = getchar()) != '\n') {
        a[charToNum(c)]++;
    }
    for (int i = 0; i < 10; ++i) {
        if (a[i] != 0)
            printf("%d:%d\n", i, a[i]);
    }
    return 0;
}

1022

将十进制整数转化为任意进制的方法：递归地先除再取余
取余的值作为该位的值，除后的值作为递归的参数。
若取余后的值大于十，即转化的进制大于十。则需要加一个if语句做判断：
if(n>=10){printf("%d",n+'A'-10);}

#include <iostream>
using namespace std;
long A, B, C;
int D;

void solve(int x) {
    long n;
    if (x) {
        solve(x / D);
        n = x % D;
        printf("%d", n);
    }
}

int main()
{
    cin >> A >> B >> D;
    C = A + B;
    if (C == 0) 
        printf("0");
    else
        solve(C);
    return 0;
}

若是要将x进制的数转化为y进制要怎么做呢？
首先将x进制转化为10进制，然后再用上面的代码将10进制转化为y进制。
x进制转化为十进制的函数：

int TenNum(char a[], int B)
{
    int len, i, num;
    int sum = 0;     
    len = strlen(a);                    //求得字符串长度
    for (i = 0; i < len; i++)
    {
        if (a[i] >= '0' && a[i] <= '9')
            num = a[i] - '0';
        else if (a[i] >= 'A' && a[i] <= 'F')
            num = a[i] - 'A' + 10;
        sum = sum * B + num;
    }
    return sum;
}

1023

#include <iostream>
#include <string.h>
using namespace std;

int main()
{
    char c;
    int a[10];
    memset(a, 0, sizeof(a));
    for (int i = 0; i < 10; ++i) {
        cin >> a[i];
    }
    for (int i = 1; i <= 9; ++i) {
        if (a[i] != 0) {
            cout << i;
            a[i]--;
            break;
        }
    }
    for (int i = 0; i <= 9; ++i) {
        while (a[i] > 0) {
            cout << i;
            a[i]--;
        }
    }
    return 0;
}

1024

....摸着测试用例过河系列

#include <iostream>
#include <string.h>
#include <string>
#define charToNum(x) (x-'0')
using namespace std;
string s;

int main()
{
    string save;
    cin >> s;
    int flag = 0;
    if (s[0] == '-')cout << '-';
    int i = 1;
    for (; s[i] != 'E'; ++i) {
        if (i != 2) {
            save += s[i];
        }
    }
    int sum = 0;
    if (s[++i] == '-')flag = 1;
    do {
        i++;
    } while (s[i] == '0');
    for (; i < s.size() ; ++i) {
        sum = sum * 10 + charToNum(s[i]);
    }
    if (flag) {
        cout << "0.";
        for (int j = 1; j < sum; ++j) {
            cout << "0";
        }
        cout << save << endl;
    }
    else {
        if (sum >= save.size()-1) {
            cout << save;
            if(sum!= save.size() - 1)
                for (int i = 0; i <= sum - save.size(); ++i)cout << "0";
        }
        else {
            int k = 0;
            for (; k <= sum; ++k) {
                cout << save[k];
            }
            cout << ".";
            for (; k < save.size(); ++k) {
                cout << save[k];
            }
        }
    }
    return 0;
}

1025

测试点5超时...我放弃了

#include <stdio.h>
#include <algorithm>
#define INF 100010
using namespace std;
struct ListNode {
    int data;
    int add;
    int nextAdd;
    int order=INF;
}List[100010];
int cnt = 0;//可能有孤立节点
int head, N, K;

bool compare(ListNode a, ListNode b) {
    return a.order < b.order;
}

int main()
{
    scanf("%d%d%d", &head, &N, &K);
    if (head == -1) {
        printf("-1\n");
        return 0;
    }
    for (int i = 0; i < N; ++i) {
        scanf("%d%d%d", &List[i].add, &List[i].data, &List[i].nextAdd);
    }
    for (int i = head; i != -1;) {
        for (int j = 0; j < N; ++j) {
            if (i == List[j].add) {
                List[j].order = cnt++;
                i = List[j].nextAdd;
                break;
            }
        }
    }
    sort(List, List+N, compare);
    for (int i = K; i <= cnt; i += K) {
        for (int j = 0; j < K / 2; ++j) {
            ListNode* temp = &List[i - K + j];
            List[i - K + j] = List[i - j - 1];
            List[i - j - 1] = *temp;
        }
    }
    for (int i = 0; i < cnt-1; ++i) {
        printf("%05d %d %05d\n", List[i].add, List[i].data, List[i + 1].add);
    }
    printf("%05d %d %d\n", List[cnt-1].add, List[cnt-1].data, -1);
    return 0;
}

1026

C语言实现浮点数四舍五入的方法：利用(int)截断的特点，将目标值+0.5即可

#include <iostream>
#include <stdio.h>
#define CLK_TCK 100
using namespace std;
int C1, C2;

void solve() {
    float t = (float)(C2 - C1)/100;
    int h = t / 3600;
    int m = (int)t % 3600 / 60;
    int s = (int)(t + 0.5) % 60;
    printf("%02d:%02d:%02d\n", h, m, s);
}

int main()
{
    cin >> C1 >> C2;
    solve();
    return 0;
}

1027

等差数列求通项与总和

#include <iostream>
#include <stdio.h>
using namespace std;
int N;
char display;
int n = 0;//行数
int rest;

void cal() {
    //通项：2*n*(n+2)+1
    while (2 * n*(n + 2) + 1 <= N)++n;
    --n;
    rest = N - 2 * n*(n + 2) - 1;
}

void show() {
    for (int i = n; i >= 0; --i) {
        for (int j = 0; j < n - i; ++j) {
            printf(" ");
        }
        for (int k = 0; k < 2 * i + 1; ++k) {
            printf("%c", display);
        }
        printf("\n");
    }
    for (int i = 1; i <= n; ++i) {
        for (int j = 0; j < n - i; ++j) {
            printf(" ");
        }
        for (int k = 0; k < 2 * i + 1; ++k) {
            printf("%c", display);
        }
        printf("\n");
    }
    printf("%d\n", rest);
}

int main()
{
    cin >> N >> display;
    cal();
    show();
    return 0;
}

1028

关于STL里的sort：如果需要自己写比较函数，容易出现错误提示invalid，这是什么原因呢？
根据文档显示，这个比较函数必须具有以下性质:

（1）两个关键字不能同时“小于等于”对方；如果k1“小于等于”k2，那么k2决不能“小于等于”k1。
（2）如果k1“小于等于”k2，且k2“小于等于”k3，那么k1必须“小于等于”k3.
（3）如果存在两个关键字，任何一个都不“小于等于”另一个，那我们称这两个关键字是“等价的”。如果k1“等价于”k2，且k2“等价于”k3，那么k1必须“等价于”k3。

所以，在重写的比较函数中，<=,>=号都不能存在，因为不满足性质一，
对于条件3，显而易见任何对于数k1，k2而言，k1<=k2和k2<=k1都是成立的，显然条件3也不符合。
对于C++的严格比较，当两数相等的时候，一定要返回false，不能返回true

#include <iostream>
#include <algorithm>
#include <string>
using namespace std;
struct pig {
    string name;
    int year;
    int month;
    int day;
}people[100010];
int N;
int cnt = 0;

bool isTrue(int year, int month, int day) {
    if (year > 1814 && year < 2014)
        return true;
    else {
        if (year == 1814)
        {
            if (month > 9)
                return true;
            else if(month==9) {
                if (day >= 6)
                    return true;
            }
        }
        if (year == 2014) {
            if (month < 9)
                return true;
            else if (month == 9) {
                if (day <= 6)
                    return true;
            }
        }
    }
    return false;
}

bool compare(pig a, pig b) {
    if (a.year > b.year)
        return true;
    else if (a.year == b.year) {
        if (a.month > b.month)
            return true;
        else if (a.month == b.month) {
            if (a.day > b.day)
                return true;
        }
    }
    return false;
}

int main()
{
    cin >> N;
    pig temp;
    for (int i = 0; i < N; ++i) {
        cin >> temp.name;
        scanf("%d/%d/%d", &temp.year, &temp.month, &temp.day);
        if (isTrue(temp.year, temp.month, temp.day)){
            people[cnt++] = temp;
        }
            
    }
    sort(people, people + cnt, compare);
    printf("%d",cnt);
    if (cnt != 0)
        cout << " " << people[0].name << " " << people[cnt - 1].name << endl;
    return 0;
}

1029

这种暴力遍历的方法时间复杂度为，但是运用哈希映射可以将时间复杂度降至

暴力遍历法：

#include <stdio.h>
#include <iostream>
#include <string>
#define toCaptital(x) (x+'A'-'a')
using namespace std;

char trans(char x) {
    if (x >= 'a'&&x <= 'z')
        return toCaptital(x);
    return x;
}

int main()
{
    string good, bad, save;
    cin >> good >> bad;
    int i = 0, j = 0;
    while (i < good.size() && j < bad.size()) {
        if (good[i] == bad[j]) {
            ++i; ++j;
        }
        else {
            save += trans(good[i++]);
        }
    }
    while (i < good.size()) {
        save += trans(good[i++]);
    }
    for (int i = 0; i < save.size() - 1; ++i) {
        for (int j = i + 1; j < save.size(); ++j) {
            if (save[i] == save[j]) {
                save[j] = ' ';
            }
        }
        if (save[i] != ' ')printf("%c", save[i]);
    }
    return 0;
}

哈希映射法：

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <string>
#define toCaptital(x) (x+'A'-'a')
using namespace std;

char trans(char x) {
    if (x >= 'a'&&x <= 'z')
        return toCaptital(x);
    return x;
}

int main()
{
    string good, bad;
    char str[100];
    int ascii[128];
    memset(ascii, 0, sizeof(ascii));
    memset(str, 0, sizeof(str));
    cin >> good >> bad;
    int cnt = 0;
    int len = 0;
    for (int i = 0; i < good.size(); ++i) {
        if (good[i] != bad[cnt]) {
            if (ascii[trans(good[i])] == 0){
                ascii[trans(good[i])]++;
                str[len++] = trans(good[i]);
            }
        }
        else {
            cnt++;
        }
    }
    printf("%s\n", str);
    return 0;
}

1030

这道题是最大子序和的变种，这种连续字序问题的解决方法都为双指针法，开始时两个指针指向数组第一个值，在满足题目条件的情况下last指针不断指向后一个值，直到不满足条件，此时first指针再不断指向后一个值直到满足条件。这样不断循环，可以得到满足的最大字序数目或最大子序和。
在最大连续子序和问题中，应该存~的和，或者使用动态规划求解。

#include <stdio.h>
#include <string.h>
#include <algorithm>
#define MAX_N 100010
using namespace std;
int N, p;
double num[MAX_N];

int main()
{
    scanf("%d %d", &N, &p);
    for (int i = 0; i < N; ++i) {
        scanf("%lf", &num[i]);
    }
    sort(num, num + N);
    int sum = 1;
    for (int first = 0, last = 0; last < N; first++) {
        while (last<N&&num[first] * p>=num[last]) {
            last++;
        }
        sum = max(sum, last - first);
    }
    printf("%d\n", sum);
    return 0;
}

1031

灵活使用hash映射让解题更方便哦！

#include<iostream>
#include <vector>
#include <string>
#define charToNum(x) (x-'0')
using namespace std;
vector<string> save(110);
char hashT[] = { '1','0' ,'X' ,'9' ,'8' ,'7' ,'6' ,'5' ,'4' ,'3' ,'2' };
int weight[] = { 7,9,10,5,8,4,2,1,6,3,7,9,10,5,8,4,2 };

bool check(string x) {
    int sum = 0;
    for (int i = 0; i < 17; ++i) {
        if (x[i] >= '0'&&x[i] <= '9')
            sum += weight[i] * charToNum(x[i]);
        else
            return false;
    }
    if (hashT[sum % 11] == x[17])
        return true;
    return false;
}

int main()
{
    int n;
    int cnt = 0;
    string pig;
    cin >> n;
    for (int i = 0; i < n; ++i) {
        cin >> pig;
        if (!check(pig)) {
            save[cnt++] = pig;
        }
    }
    if (cnt == 0)
        cout << "All passed" << endl;
    else {
        for (int i = 0; i < cnt; ++i) {
            cout << save[i] << endl;
        }
    }
    return 0;
}

1032

#include <stdio.h>
#include <algorithm>
#include <string.h>
#define MAX_N 100010
using namespace std;
int N;
int HashT[MAX_N];

int main()
{
    int index, score;
    memset(HashT, 0, sizeof(HashT));
    scanf("%d", &N);
    for (int i = 0; i < N; ++i) {
        scanf("%d %d", &index, &score);
        HashT[index] += score;
    }
    int maxN = 0;
    for (int i = 1; i <= N; ++i) {
        if (HashT[i] >= HashT[maxN]) maxN = i;
    }
    printf("%d %d\n", maxN, HashT[maxN]);
    return 0;
}

1033

#include <stdio.h>
#include <string.h>
#define toSmall(x) (x-'A'+'a')
#define MAX_N 100010
int isOK[128];
char output[MAX_N];

bool check(char c) {
    if (isOK[c])
        return false;
    if (isOK['+'] && c >= 'A'&&c <= 'Z')
        return false;
    return true;
}

int main()
{
    memset(isOK, 0, sizeof(isOK));
    char c;
    int cnt = 0;
    while ((c = getchar()) != '\n') { 
        isOK[c] = 1;
        if (c >= 'A'&&c <= 'Z')
            isOK[toSmall(c)] = 1;
    }
    while ((c = getchar()) != '\n') {
        if (check(c)) output[cnt++] = c;
    }
    printf("%s\n", output);
    return 0;
}

1034

是一道很水但是很麻烦的题。
题目描述中说分子分母输入的范围不超过int的范围，但是两个数相乘有可能超过int的范围，所以必须使用long long
另外复习了一下辗转相除法gcd：
用较大数除以较小数，若余数为0，则较小数为最大公约数
若余数不为0，用较小数除以余数

#include <stdio.h>
#include <math.h>
#include <iostream>
using namespace std;
struct num {
    long long inter;
    long long son;
    long long mom;
};

int gcd(long son, long mom) {
    if (son == 0)return mom;
    return gcd(mom%son, son);
}

void show(num num1) {
    if (num1.mom == 0) {
        printf("Inf");
        return;
    }
    bool flag = false;
    if (num1.mom < 0) {
        num1.son = -num1.son;
        num1.mom = abs(num1.mom);
    }
    if (num1.son < 0) {
        flag = true;
        num1.son = abs(num1.son);
    }
    num1.inter = num1.son / num1.mom;
    num1.son %= num1.mom;
    int p = gcd(num1.son, num1.mom);
    num1.son /= p;
    num1.mom /= p;
    if (flag)printf("(-");
    if (num1.inter&&num1.son&&num1.mom) {
        printf("%lld %lld/%lld", num1.inter, num1.son, num1.mom);
    }
    else if (num1.inter==0&&num1.son&&num1.mom) {
        printf("%lld/%lld",num1.son, num1.mom);
    }
    else if (num1.son == 0) {
        printf("%lld", num1.inter);
    }
    if (flag)printf(")");
}

void display(num num1, num num2, num num3,char x) {
    show(num1);
    printf(" %c ", x);
    show(num2);
    printf(" = ");
    show(num3);
    printf("\n");
}

void add(num num1, num num2) {
    num pig;
    pig.mom = num1.mom*num2.mom;
    pig.son = num1.son*num2.mom + num2.son*num1.mom;
    display(num1, num2, pig, '+');
}

void dele(num num1, num num2) {
    num pig;
    pig.mom = num1.mom*num2.mom;
    pig.son = num1.son*num2.mom - num2.son*num1.mom;
    display(num1, num2, pig, '-');
}

void mul(num num1,num num2) {
    num pig;
    pig.mom = num1.mom*num2.mom;
    pig.son = num1.son*num2.son;
    display(num1, num2, pig, '*');
}

void chu(num num1, num num2) {
    num pig;
    pig.mom = num1.mom*num2.son;
    pig.son = num1.son*num2.mom;
    display(num1, num2, pig, '/');
}

int main()
{
    num num1, num2;
    scanf("%lld/%lld %lld/%lld", &num1.son, &num1.mom, &num2.son, &num2.mom);
    add(num1, num2);
    dele(num1, num2);
    mul(num1, num2);
    chu(num1, num2);
    return 0;
}

1035

#include <iostream>
#include <algorithm>
#include <stdio.h>
#define MAX_N 110
using namespace std;
int arr1[MAX_N], arr2[MAX_N];
int N;

int check() {
    int i;
    for (i = 1; i < N; ++i) {
        if (arr2[i - 1] > arr2[i])
            break;
    }
    int j;
    for (j = i; j < N; ++j) {
        if (arr1[j] != arr2[j])
            break;
    }
    if (j == N)
        return i;
    return -1;
}

void merge_arr(int k) {
    int i;
    for (i = k; i <= N; i += k) {
        sort(arr2 + i - k, arr2 + i);
    }
    sort(arr2 + i - k, arr2 + N);
}

void solve() {
    int t;
    if ((t = check()) != -1) {
        cout << "Insertion Sort" << endl;
        sort(arr2, arr2 + t + 1);
    }
    else {
        cout << "Merge Sort" << endl;
        int flag;
        for (int l = 2; l < N; l *= 2) {
            for (int k = l; k < N; k += 2l) {
                if (arr2[k] < arr2[k - 1]) {
                    flag = l * 2;
                    goto pig;
                }
            }
        }
        pig:
            merge_arr(flag);
    }
    printf("%d", arr2[0]);
    for (int i = 1; i < N; ++i) {
        printf(" %d", arr2[i]);
    }
}

int main()
{
    scanf("%d", &N);
    for (int i = 0; i < N; ++i) {
        scanf("%d", &arr1[i]);
    }
    for (int i = 0; i < N; ++i) {
        scanf("%d", &arr2[i]);
    }
    solve();
    return 0;
}

1036

怎么说呢...送分题

#include <stdio.h>
using namespace std;

int main()
{
    float N;
    char x;
    scanf("%f %c", &N, &x);
    int Y = (int)(N / 2 + 0.5);
    for (int i = 0; i < N; ++i) {
        printf("%c", x);
    }
    printf("\n");
    for (int i = 0; i < Y - 2; ++i) {
        printf("%c", x);
        for (int j = 0; j < N - 2; ++j) {
            printf(" ");
        }
        printf("%c\n", x);
    }
    for (int i = 0; i < N; ++i) {
        printf("%c", x);
    }
    return 0;
}

1037

这道题有点进制转换的意思...主要思路就是先把所有的钱换到最低位再相减，得到的余数再进行进位，这样的话比较便于操作

#include <stdio.h>
#include <math.h>
using namespace std;
struct money {
    int G;
    int S;
    int K;
};

int trans(money temp) {
    return temp.G * 17 * 29 + temp.S * 29 + temp.K;
}

int main()
{
    money shouldP, P, R;
    scanf("%d.%d.%d %d.%d.%d", &shouldP.G, &shouldP.S, &shouldP.K, &P.G, &P.S, &P.K);
    R.K = trans(P) - trans(shouldP);
    R.S = R.K / 29;
    R.K = abs(R.K % 29);
    R.G = R.S / 17;
    R.S = abs(R.S % 17);
    printf("%d.%d.%d\n", R.G, R.S, R.K);
    return 0;
}

1038

这道题存学生的分数当然可以定义一个足够大的数组，一个学生一个学生的放，这样的话查找的时间复杂度为
但是若用一个哈希映射表，不仅省下了空间，而且查找的时间复杂度变成了

#include <stdio.h>
#include <math.h>
#define MAX_N 100010
using namespace std;
int N, K;
int stu[101];
int save[MAX_N];

int main()
{
    int j;
    scanf("%d", &N);
    for (int i = 0; i < N; ++i) {
        scanf("%d", &j);
        ++stu[j];
    }
    int k;
    scanf("%d", &K);
    for (int i = 0; i < K; ++i) {
        scanf("%d", &k);
        save[i] = stu[k];
    }
    printf("%d", save[0]);
    for (int i = 1; i < K; ++i) {
        printf(" %d", save[i]);
    }
    return 0;
}

1039

此题与坏了键盘那道题很像，同样有两种方法，一种是暴力遍历法，一种是哈希映射法。但是此题用遍历法也非常高效，前提是先将字符串按ascii顺序排序。
遍历法：

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
string buy, me;

int main()
{
    cin >> buy >> me;
    sort(buy.begin(), buy.end());
    sort(me.begin(), me.end());
    int cnt = 0;
    int j = 0;
    for (int i = 0; i < buy.size(); ++i) {
        while (buy[i] > me[j] && j < me.size()) {
            ++j;
        }
        if (buy[i] == me[j] && j < me.size()) { 
            ++cnt; 
            ++j;
        }
    }
    if (cnt == me.size()) {
        cout << "Yes " << buy.size() - cnt << endl;
    }
    else {
        cout << "No " << me.size() - cnt << endl;
    }
    return 0;
}

哈希映射：

#include <stdio.h>
#include <string.h>

int main()
{
    char c;
    int ascii[128];
    memset(ascii, 0, sizeof(ascii));
    while ((c = getchar()) != '\n') ++ascii[c];
    while ((c = getchar()) != '\n') --ascii[c];//若ascii数组中存在某个数小于0，则输出No，反之输出Yes
    bool flag = true;
    int sumY = 0, sumN = 0;
    for (int i = 0; i < 128; ++i) {
        if (ascii[i] < 0) {
            flag = false;
            sumN += -ascii[i];
        }
        if (flag) {
            sumY += ascii[i];
        }
    }
    flag ? printf("Yes %d\n", sumY) : printf("No %d\n", sumN);
    return 0;
}

1040

这道题就是一个简单的数学问题，对每一个A，构造出的PAT个数等于A之前P的个数x乘以A之后T的个数y，令z=x*y，即求

#include <stdio.h>
#include <string.h>
#define MAX_N 100010
char save[MAX_N];
int num[3] = { 0 };
long sum = 0;

int main()
{
    char c;
    int cnt = 0;
    while ((c = getchar()) != '\n') {
        save[cnt++] = c;
        if (c == 'P')++num[0];
        else if (c == 'A')++num[1];
        else ++num[2];
    }
    int numP = 0;
    int numT = num[2];
    for (int i = 0; i < cnt; ++i) {
        if (save[i] == 'P')numP++;
        else if (save[i] == 'T')numT--;
        else if (save[i] == 'A')sum += numP * numT;
        sum %= 1000000007;
    }
    printf("%d\n", sum);
    return 0;
}

1041

因为N<1000，数据量很小，直接遍历即可

#include <stdio.h>
struct sit {
    char id[14];
    int test_sit;
    int exam_sit;
}sits[1010];


int main()
{
    int finds[1010];
    int N, M, find;
    scanf("%d", &N);
    for (int i = 0; i < N; ++i) {
        scanf("%s %d %d", &sits[i].id, &sits[i].test_sit, &sits[i].exam_sit);
    }
    scanf("%d", &M);
    for (int i = 0; i < M; ++i) {
        scanf("%d", &find);
        for (int j = 0; j < N; ++j) {
            if (find == sits[j].test_sit)finds[i] = j;
        }
    }
    for (int i = 0; i < M; ++i) {
        printf("%s %d\n", sits[finds[i]].id, sits[finds[i]].exam_sit);
    }
    return 0;
}

1042

天啊这种题都要写吐了

#define upToLow(x) (x-'A'+'a')
#include <stdio.h>
#include <string.h>
#include <algorithm>


int main()
{
    int ASCII[128];
    char c;
    memset(ASCII, 0, sizeof(ASCII));
    while ((c = getchar()) != '\n') {
        if (c >= 'a'&&c <= 'z')++ASCII[c];
        else if (c >= 'A'&&c <= 'Z')++ASCII[upToLow(c)];
    }
    int maxN = 0;
    for (int i = 1; i < 128; ++i) {
        ASCII[maxN] < ASCII[i] ? maxN = i : maxN;
    }
    printf("%c %d\n", maxN, ASCII[maxN]);
    return 0;
}

1043

反正才六个字母，就不用写循环了

#include <stdio.h>
#include <string.h>
#include <algorithm>

int main()
{
    char c;
    int save[128] = { 0 };
    while ((c = getchar()) != '\n') {
        ++save[c];
    }
    while (save['P'] || save['A'] || save['T'] || save['e'] || save['s'] || save['t']) {
        if (save['P']) { printf("P"); save['P']--; }
        if (save['A']) { printf("A"); save['A']--; }
        if (save['T']) { printf("T"); save['T']--; }
        if (save['e']) { printf("e"); save['e']--; }
        if (save['s']) { printf("s"); save['s']--; }
        if (save['t']) { printf("t"); save['t']--; }
    }
    return 0;
}

1044

也没觉得特别难...就是写着写着心态崩了。
打表还是太浪费时间了....
这道题不管是scanf还是cin都不好用，最好使用geiline
getline有三个参数，第一个是输入流cin，第二个是接收输入的字符串，第三个默认是换行。

#include <stdio.h>
#include <string>
#include <vector>
#include <map>
#include <iostream>
#include <algorithm>
#define charToNum(x) (x-'0')
using namespace std;
struct num {
    int ge;
    int shi;
    int bai;
};
struct fire {
    int G;
    int S;
};
int N;
const char* fireG[] = { "tret","jan", "feb", "mar", "apr", "may", "jun", "jly", "aug", "sep", "oct", "nov", "dec" };
const char* fireS[] = { "","tam", "hel", "maa", "huh", "tou", "kes", "hei", "elo", "syy", "lok", "mer", "jou" };
map<string, int> hashT;

num transN(string temp) {
    num x;
    if (temp.size() == 3) {
        x.bai = charToNum(temp[0]);
        x.shi = charToNum(temp[1]);
        x.ge = charToNum(temp[2]);
    }
    else if (temp.size() == 2) {
        x.shi = charToNum(temp[0]);
        x.ge = charToNum(temp[1]);
        x.bai = 0;
    }
    else {
        x.ge = charToNum(temp[0]);
        x.shi = 0;
        x.bai = 0;
    }
    return x;
}
int main()
{
    hashT["tret"] = 0; hashT["jan"] = 1; hashT["feb"] = 2; hashT["mar"] = 3; hashT["apr"] = 4; hashT["may"] = 5;
    hashT["jun"] = 6; hashT["jly"] = 7; hashT["aug"] = 8; hashT["sep"] = 9; hashT["oct"] = 10; hashT["nov"] = 11;
    hashT["dec"] = 12; hashT["tam"] = 13; hashT["hel"] = 26; hashT["maa"] = 39; hashT["huh"] = 52; hashT["tou"] = 65;
    hashT["kes"] = 78; hashT["hei"] = 91; hashT["elo"] = 104; hashT["syy"] = 117; hashT["lok"] = 130; hashT["mer"] = 143;
    hashT["jou"] = 156;
    scanf("%d", &N);
    getchar();
    char c;
    vector<string> save(N);
    string pig;
    for (int i = 0; i < N; ++i) {
        getline(cin, save[i]);
    }
    for (int i = 0; i < N; ++i) {
        if (save[i][0] >= '0'&&save[i][0] <= '9') {
            num x = transN(save[i]);
            int all = x.bai * 100 + x.shi * 10 + x.ge;
            fire y;
            y.G = all % 13;
            y.S = all / 13;
            if (y.S) {
                if (y.G != 0)
                    printf("%s %s\n", fireS[y.S], fireG[y.G]);
                else
                    printf("%s\n", fireS[y.S]);
            }
            else {
                printf("%s\n", fireG[y.G]);
            }
        }
        else {
            int sum = 0;
            string temp;
            for (int j = 0; j < save[i].size(); ++j) {
                if (save[i][j] == ' ') {
                    sum += hashT[temp];
                    temp = "";
                }
                else temp += save[i][j];
            }
            sum += hashT[temp];
            cout << sum << endl;
        }
    }
    return 0;
}

1045

这道题的方法还是很好想的，只需遍历两次数组即可。
第一次从左向右遍历，不断更新最大值
第二次从右向左遍历，不断更新最小值

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <algorithm>
#define Max_N 100010
using namespace std;

int main()
{
    long N, maxN = 0, minN = 1000000001;
    long a[Max_N];
    bool isOK[Max_N];
    memset(isOK, 1, sizeof(isOK));
    scanf("%d", &N);
    int cnt = N;
    for (int i = 0; i < N; ++i) {
        scanf("%ld", &a[i]);
        if (maxN < a[i]) {
            maxN = a[i];
        }
        else {
            isOK[i] = false;
            cnt--;
        }
    }
    for (int j = N - 1; j >= 0; --j) {
        if (minN > a[j]) {
            minN = a[j];
        }
        else {
            if (isOK[j]) {
                isOK[j] = false;
                cnt--;
            }
        }
    }
    printf("%ld\n", cnt);
    bool flag = true;
    for (int i = 0; i < N; ++i) {
        if (isOK[i]){
            if (flag) {
                printf("%ld", a[i]);
                flag = false;
            }
            else {
                printf(" %ld", a[i]);
            }
        }
    }
    putchar('\n');
    return 0;
}

1046

#include <stdio.h>
#include <iostream>
using namespace std;

int main()
{
    int N, a1, a2, b1, b2;
    int cnt_a = 0, cnt_b = 0;
    scanf("%d", &N);
    for (int i = 0; i < N; ++i) {
        scanf("%d %d %d %d", &a1, &a2, &b1, &b2);
        if (a1 + b1 == a2 && a1 + b1 != b2) {
            cnt_b++;
        }
        else if (a1 + b1 == b2 && a1 + b1 != a2) {
            cnt_a++;
        }
    }
    printf("%d %d\n", cnt_a, cnt_b);
    return 0;
}

1047

#include <stdio.h>
#include <iostream>
#include <algorithm>
using namespace std;
struct que {
    int grade = 0;
    int id;
}all[1001];


int main()
{
    int N;
    scanf("%d", &N);
    for (int i = 0; i < N; ++i) {
        int temp_que, temp_id, temp_grade;
        scanf("%d-%d %d", &temp_que, &temp_id, &temp_grade);
        all[temp_que].id = temp_que;
        all[temp_que].grade += temp_grade;
    }
    int max_grade = 0, max_id;
    for (int i = 0; i < 1000; ++i) {
        if (all[i].grade > max_grade) {
            max_grade = all[i].grade;
            max_id = all[i].id;
        }
    }
    printf("%d %d\n", max_id, max_grade);
}

1048

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <iostream>
#define charToNum(x) (x-'0')
using namespace std;
char A[100], B[100], solve[100];
char trans[] = { '0','1','2','3','4','5','6','7','8','9','J','Q','K' };


int main()
{
    scanf("%s %s", A, B);
    int len_A = strlen(A);
    int len_B = strlen(B);
    int len_min, len_max;
    char choose;
    if (len_A > len_B) {
        len_max = len_A; len_min = len_B; choose = 'A';
    }
    else { len_min = len_A; len_max = len_B; choose = 'B'; }
    int distance = len_max - len_min;
    bool flag;
    len_max % 2 == 0 ? flag = false : flag = true;
    for (int i = 0; i < distance; ++i) {
        if (choose == 'A') {
            if (flag) {
                solve[i] = A[i];
            }
            else {
                int sum = -charToNum(A[i]);
                sum < 0 ? sum += 10 : sum;
                solve[i] = trans[sum];
            }
        }
        else {
            solve[i] = B[i];
        }
        flag = !flag;
    }
    for (int i = 0; i < len_min; ++i) {
        int j = i, k = i;
        if (choose == 'A')
            j = distance + i;
        else
            k = distance + i;
        if (flag) {
            int sum = (charToNum(A[j]) + charToNum(B[k])) % 13;
            solve[i+distance] = trans[sum];
        }
        else {
            int sum = charToNum(B[k]) - charToNum(A[j]);
            sum < 0 ? sum += 10 : sum;
            solve[i+distance] = trans[sum];
        }
        flag = !flag;
    }
    printf("%s\n",solve);
}

1049

#include <stdio.h>

int main()
{
    int N;
    double ai, sum = 0;
    scanf("%d", &N);
    for (int i = 0; i < N; ++i) {
        scanf("%lf", &ai);
        sum += ai * (N - i)*(i + 1);
    }
    printf("%.2lf\n", sum);
}

1050

#include <stdio.h>
#include <string.h>
#include <algorithm>
int N, m, n;
int save[10010];
int matrix[10010][10010] = { 0 };

bool compare(int a, int b) {
    return a > b;
}

void show() {
    for (int i = 0; i < m; ++i) {
        for (int j = 0; j < n; ++j) {
            if (j == 0)printf("%d", matrix[j][i]);
            else printf(" %d", matrix[j][i]);
        }
        putchar('\n');
    }
}

void solve() {
    int cnt = 0;
    matrix[0][0] = save[0];
    int x = 0, y = 0;
    while (cnt < m*n - 1) {
        while (x + 1 < n && !matrix[x + 1][y]) {
            matrix[++x][y] = save[++cnt];
        }//first
        while (y + 1 < m && !matrix[x][y + 1]) {
            matrix[x][++y] = save[++cnt];
        }//second
        while (x - 1 >= 0 && !matrix[x - 1][y]) {
            matrix[--x][y] = save[++cnt];
        }//third
        while (y - 1 >= 0 && !matrix[x][y - 1]) {
            matrix[x][--y] = save[++cnt];
        }//forth
    }
    show();
}


int main()
{
    scanf("%d", &N);
    for (int i = 0; i < N; ++i) {
        scanf("%d", &save[i]);
    }
    for (int i = 1; i*i <= N; ++i) {
        if (N%i == 0) {
            n = i;
        }
    }
    m = N / n;
    sort(save, save + N, compare);
    solve();
}

1051

这题真的相当的迷...
用float是无法通过的，必须用double
需要注意的是，浮点数相加减需要用<EOF判断是否为0，而不可以直接判断是否为0。.2lf的截断也不太常规，一个非常接近0的负数截断之后会变为，所以小于EOF的数要在判断前置0。
printf也有好用的地方，当%后加+号后，可以显示带符号的数。
如果%后加空格，如果值为正，显示空格但不显示＋，若为负，则显示-
补充一下，为浮点类型的数求绝对值，为整型求绝对值

#include <stdio.h>
#include <math.h>
#define EOF 0.01

int main()
{
    double R1,P1,R2,P2;
    scanf("%lf %lf %lf %lf",&R1,&P1,&R2,&P2);
    double result_A=R1*cos(P1)*R2*cos(P2)-R1*sin(P1)*R2*sin(P2);
    double result_B=R1*cos(P1)*R2*sin(P2)+R1*sin(P1)*R2*cos(P2);
    if(fabs(result_A)<EOF)result_A=0;
    if(fabs(result_B)<EOF)result_B=0;
    printf("%.2lf%+.2lfi",result_A,result_B);
    return 0;
}

1052

这是什么大臭题啊哭了。。。
%[^]]表示输入一个字符串，以]结束，相当于%s的增强版。

#include <stdio.h>

int main()
{
    int N,input[10000][5];
    char bq[3][10][5],c;
    int cnt[3]={0};
    for(int i=0;i<3;++i){
        for(int index=0;(c=getchar())!='\n';){
            if(c=='['){
                scanf("%[^]]",bq[i][index++]);
                cnt[i]++;
            }
        }
    }
    scanf("%d",&N);
    for(int i=0;i<N;++i){
        for(int j=0;j<5;++j){
            scanf("%d",&input[i][j]);
            if((j==0||j==4)&&(input[i][j]>cnt[0]||input[i][j]<1))input[i][0]=-1;
            else if((j==1||j==3)&&(input[i][j]>cnt[1]||input[i][j]<1))input[i][0]=-1;
            else if((j==2)&&(input[i][j]>cnt[2]||input[i][j]<1))input[i][0]=-1;
        }
    }
    for(int i=0;i<N;++i){
        if(input[i][0]==-1){
            puts("Are you kidding me? @\\/@");
        }else
        {
            printf("%s(%s%s%s)%s\n",bq[0][input[i][0]-1],bq[1][input[i][1]-1],bq[2][input[i][2]-1],bq[1][input[i][3]-1],bq[0][input[i][4]-1]);
        }
    }
    return 0;
}

1053

printf中打出%号需输入%%

#include <stdio.h>

int main()
{
    int N,D,cnt_m=0,cnt_n=0;
    double e,elec;
    scanf("%d %lf %d",&N,&e,&D);
    int cnt,days;
    for (int i = 0; i < N; i++)
    {
        cnt=0;
        scanf("%d",&days);
        for(int j=0;j<days;++j){
            scanf("%lf",&elec);
            if(elec<e)++cnt;
        }
        if(cnt*2>days){
            days>D?++cnt_n:++cnt_m;
        }
    }
    double m=cnt_m*100/(double)N;
    double n=cnt_n*100/(double)N;
    printf("%.1lf%% %.1lf%%",m,n);
    return 0;
}

1054

了解了sscanf和sprintf的神奇用法
sscanf：读取的字符串，读取的格式，写入的对象
sprintf：写入的对象，写入的格式，读取的对象
注意哦，puts自带空格，不用单独加了。

#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;

int main()
{
    char a[50],b[50];
    double sum=0,temp;
    int N,cnt=0;
    cin>>N;
    for(int i=0;i<N;++i){
        scanf("%s",a);
        sscanf(a,"%lf",&temp);
        sprintf(b,"%.2f",temp);
        bool flag=true;
        for(int j=0;j<strlen(a);++j){
            if(a[j]!=b[j]){flag=false;break;}
        }
        if(!flag||temp<-1000||temp>1000){
            printf("ERROR: %s is not a legal number\n",a);
            continue;
        }else{
            sum+=temp;
            cnt++;
        }
    }
    if(cnt==0){
        puts("The average of 0 numbers is Undefined");
    }else if(cnt==1){
        printf("The average of 1 number is %.2f",sum);
    }else{
        printf("The average of %d numbers is %.2f",cnt,sum/cnt);
    }
    return 0;
}

1055

strcmp:
设这两个字符串为str1，str2，
若str1=str2，则返回零；
若str1<str2，则返回负数；
若str1>str2，则返回正数。

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
struct people{
    int tall;
    char name[10];
}queue[10010];
int N,K;
people arrange[10010];

bool compare(people a,people b){
    return a.tall!=b.tall?a.tall<b.tall:strcmp(a.name,b.name)>0;
}

int main()
{
    scanf("%d %d",&N,&K);
    for(int i=0;i<N;++i){
        scanf("%s %d",&queue[i].name,&queue[i].tall);
    }
    sort(queue,queue+N,compare);
    int step=N/K,cnt=0;
    int last_step=N-step*(K-1);
    for(int i=0;i<N-last_step;i+=step){
        for(int j=i+step-1;j>=i;--j){
            int pos=i+step/2+1+cnt;
            cnt<0?cnt=-cnt:cnt=-cnt-1;
            arrange[pos]=queue[j];
        }
        cnt=0;
    }
    for(int i=N-1;i>=N-last_step;--i){
        int pos=last_step/2+1+cnt+N-last_step;
        cnt<0?cnt=-cnt:cnt=-cnt-1;
        arrange[pos]=queue[i];
    }
    bool flag=true;
    for(int i=N-last_step+1;i<=N;++i){
        if(flag){
            printf("%s",arrange[i].name);
            flag=false;
        }else{
            printf(" %s",arrange[i].name);
        }
    }
    putchar('\n');
    for(int i=N-last_step-step;i>=0;i-=step){
        flag=true;
        for(int j=i+1;j<=i+step;j++){
            if(flag){
                printf("%s",arrange[j].name);
                flag=false;
            }else{
                printf(" %s",arrange[j].name);
            }
        }
        putchar('\n');
    }
    return 0;
}

1056

数学问题，推出通式为

#include <iostream>
using namespace std;

int main()
{
    int N,sum=0,temp;
    cin>>N;
    for(int i=0;i<N;++i){
        cin>>temp;
        sum+=temp*11*(N-1);
    }
    cout<<sum<<endl;
    return 0;
}

1057

进制转换问题，思路挺简单的

#include <iostream>
#define upperToNum(x) (x-'A')
#define lowerToNum(x) (x-'a')
using namespace std;
int sum=0,cnt_0=0,cnt_1=0;

void solve(){
    if(sum==0)
        return;
    if(sum%2==0)cnt_0++;
    else cnt_1++;
    sum/=2;
    solve();
}

int main()
{
    char c;
    
    while((c=getchar())!='\n'){
        if(c>='A'&&c<='Z')sum+=upperToNum(c)+1;
        else if(c>='a'&&c<='z')sum+=lowerToNum(c)+1;
    }
    solve();
    cout<<cnt_0<<" "<<cnt_1<<endl;
    return 0;
}

1058

令人害怕的大模拟题...除了硬着头皮往下写也没什么其他办法

#define charToNum(x) (x-'0')
#include <iostream>
#include <algorithm>
#include <string>
using namespace std;
struct prob{
    int grades;
    int all_choose;
    int cor_choose;
    int wrong;
    string save;
}probs[100];
int N,M;


int main()
{
    cin>>N>>M;
    char c;
    for(int i=0;i<M;++i){
        probs[i].wrong=0;
        cin>>probs[i].grades>>probs[i].all_choose>>probs[i].cor_choose;
        for(int j=0;j<probs[i].cor_choose;++j){
            cin>>c;
            probs[i].save+=c;
        }
    }
    string person;
    int max_wrong=0;
    for(int i=0;i<N;++i){
        int num=-1,cnt=0,sum=0;
        string compare;
        getchar();
        getline(cin,person);
        for(int i=0;i<person.size();++i){
            if(person[i]>='0'&&person[i]<='9')num=charToNum(person[i]);
            if(person[i]>='a'&&person[i]<='z')compare+=person[i];
            if(person[i]==')'){
                sort(compare.begin(),compare.end());
                if(num==probs[cnt].cor_choose&&compare==probs[cnt].save)
                    sum+=probs[cnt].grades;
                else probs[cnt].wrong++;
                compare="";
                cnt++;
                max_wrong=max(probs[cnt].wrong,max_wrong);
            }
        }
        cout<<sum<<endl;
    }
    if(max_wrong==0){
        cout<<"Too simple"<<endl;
    }else{
        cout<<max_wrong;
        for(int i=0;i<M;++i){
            if(probs[i].wrong==max_wrong)
                cout<<" "<<i+1;
        }
    }
    return 0;
}

1059

这题用哈希散列更加方便而且大大降低了复杂度，用空间换时间

#include <vector>
#include <string>
#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
int N,M;
int people[10010];
bool visit[10010];

bool isPrime(int x){
    for(int i=2;i*i<=x;++i){
        if(x%i==0)
            return false;
    }
    return true;
}

void finding(int find){
    if(visit[find]){
        printf("%04d: Checked\n",find);
        return;
    }
    else if(people[find]==0){
        printf("%04d: Are you kidding?\n",find);
        return;
    }
    if(people[find]==1){
        printf("%04d: Mystery Award\n",find);
        visit[find]=true;
        return;
    }else if(isPrime(people[find])){
        printf("%04d: Minion\n",find);
        visit[find]=true;
        return;
    }else{
        printf("%04d: Chocolate\n",find);
        visit[find]=true;
        return;
    }
}

int main()
{
    memset(visit,0,sizeof(visit));
    memset(people,0,sizeof(people));
    int num;
    scanf("%d",&N);
    for(int i=0;i<N;++i){
        scanf("%d",&num);
        people[num]=i+1;
    }
    int find;
    scanf("%d",&M);
    for(int i=0;i<M;++i){
        scanf("%d",&find);
        finding(find);
    }
    return 0;
}

1060

这道题其实并没有卡时间，但是保险起见我还是用了二分法。
这是一道极大最小值的模板题。

#include <stdio.h>
#include <iostream>
#include <algorithm>
using namespace std;
int N,save[100010];

void solve(){
    int left=1,right=N,mid;
    while(right-left>1){
        mid=(left+right)/2;
        if(save[mid]>mid)left=mid;
        else right=mid;
    }
    save[right]>right?mid=right:mid=left;
    printf("%d\n",mid);
}

bool compare(int a,int b){
    return a>b;
}

int main()
{
    int max_num=0;
    scanf("%d",&N);
    for(int i=1;i<=N;++i){
        scanf("%d",&save[i]);
        max_num=max(max_num,save[i]);
    }
    if(max_num==1)
    {
        printf("0\n");
        return 0;
    }
    sort(save+1,save+N+1,compare);
    solve();
    return 0;
}

1061

#include <stdio.h>
#include <iostream>
using namespace std;
struct prob{
    int grade;
    bool ans;
}probs[110];

int main()
{
    int N,M;
    cin>>N>>M;
    for(int i=0;i<M;++i){
        cin>>probs[i].grade;
    }
    for(int i=0;i<M;++i){
        cin>>probs[i].ans;
    }
    for(int i=0;i<N;++i){
        int sum=0;
        bool temp;
        for(int i=0;i<M;++i){
            cin>>temp;
            if(temp==probs[i].ans)
                sum+=probs[i].grade;
        }
        cout<<sum<<endl;
    }
    return 0;
}

1062

主要是读题时间比较久，姥姥要好好提高语文水平呀

#include <stdio.h>
#include <iostream>
using namespace std;

int gcd(int a,int b){
    if(b==0)return a;
    return gcd(b,a%b);
}

void swap(int &a,int &b){
    int temp=a;
    a=b;
    b=temp;
}

int main()
{
    int son1,son2,mom1,mom2,mom;
    scanf("%d/%d %d/%d %d",&son1,&mom1,&son2,&mom2,&mom);
    double x1=(double)son1/(double)mom1;
    double x2=(double)son2/(double)mom2;
    if(x1>x2) swap(x1,x2);
    bool flag=true;
    for(int i=0;i<1000;++i){
        if(gcd(mom,i)!=1)continue;
        double temp=(double)i/(double)mom;
        if(temp>x1&&temp<x2){
            if(flag){
                printf("%d/%d",i,mom);
                flag=false;
            }else{
                printf(" %d/%d",i,mom);
            }
        }
        if(temp>=x2)break;
    }
    return 0;
}

1063

emmmm

#include <stdio.h>
#include <algorithm>
#include <math.h>
#include <iostream>
using namespace std;


int main()
{
    int N,n,v;
    cin>>N;
    int max_n=0;
    for(int i=0;i<N;++i){
        cin>>n>>v;
        max_n=max(max_n,n*n+v*v);
    }
    printf("%.2f\n",sqrt(max_n));
}

1064

set大法好，柳神牛逼！

#include <stdio.h>
#include <set>
#include <iostream>
#define charToInt(x) (x-'0')
using namespace std;


int main()
{
    char c;
    int sum=0,N;
    set<int> save;
    cin>>N;getchar();
    while((c=getchar())!='\n'){
        if(c==' '){
            save.insert(sum);
            sum=0;
        }else{
            sum+=charToInt(c);
        }
    }
    save.insert(sum);
    set<int>::iterator scan=save.begin();
    bool flag=true;
    cout<<save.size()<<endl;
    while(scan!=save.end()){
        if(flag){
            cout<<*scan;
            flag=false;
        }else{
            cout<<" "<<*scan;
        }
        scan++;
    }
}

1065

因为是的复杂度，把大数组全部遍历一遍也问题不大

#include <stdio.h>
#include <set>
#include <string.h>
#include <iostream>
#define MAX_N 100010
using namespace std;
bool people[MAX_N];
int couple[MAX_N];

int main()
{
    int N,M,id1,id2,id;
    set<int> save;
    set<int>::iterator scan;
    memset(people,0,sizeof(people));
    memset(couple,-1,sizeof(couple));
    scanf("%d",&N);
    for(int i=0;i<N;++i){
        scanf("%d %d",&id1,&id2);
        couple[id1]=id2;
        couple[id2]=id1;
    }
    scanf("%d",&M);
    for(int i=0;i<M;++i){
        scanf("%d",&id);
        people[id]=true;
    }
    for(int i=0;i<MAX_N;++i){
        if(people[i]&&(couple[i]==-1||(couple[i]!=-1&&people[couple[i]]==false))){
            save.insert(i);
        }
    }
    printf("%d\n",save.size());
    bool flag=true;
    for(scan=save.begin();scan!=save.end();scan++){
        if(flag){
            printf("%05d",*scan);
            flag=false;
        }else{
            printf(" %05d",*scan);
        }
    }
}

1066

#include <stdio.h>
#include <iostream>
using namespace std;
int colors[500][500];

int main()
{
    int M,N,A,B,color;
    scanf("%d %d %d %d %d",&M,&N,&A,&B,&color);
    for(int i=0;i<M;++i)
        for(int j=0;j<N;++j){
            scanf("%d",&colors[i][j]);
            if(colors[i][j]>=A&&colors[i][j]<=B)colors[i][j]=color;
        }
    for(int i=0;i<M;++i){
        for(int j=0;j<N;++j){
            j==0?printf("%03d",colors[i][j]):printf(" %03d",colors[i][j]);
        }
        putchar('\n');
    }
    return 0;
}

1067

这题只是正确的密码不包含空格，没说用户输入的密码不包含空格，所以scanf或者cin都不能用，只能用getline。

#include <iostream>
#include <string>
using namespace std;


int main()
{
    int N,cnt=0;
    string psw;
    cin>>psw>>N;
    string input;
    getchar();
    while(1){
        getline(cin,input);
        if(input=="#")
            break;
        if(input==psw){
            cout<<"Welcome in";
            break;
        }
        else{
            cout<<"Wrong password: "<<input<<endl;
        }
        if(++cnt==N){
            cout<<"Account locked";
            break;
        }
    }
    return 0;
}

1068

基础的搜索问题
为啥用int可以ac，用long就有一个测试点答案错误？

#include <stdio.h>
#include <iostream>
#include <math.h>
#include <map>
#include <string.h>
using namespace std;
int canvas[1000][1000];
map<int,int> flag;
int step;
int M,N;

bool examine(int x,int y){
    for(int i=-1;i<=1;++i)
        for(int j=-1;j<=1;++j){
            if((i==0&&j==0)||(x+i<0)||(x+i>=N)||(y+j<0)||(y+j>=M))
                continue;
            else if(abs(canvas[x][y]-canvas[x+i][y+j])<=step){
                return false;
            }
        }
    return true;
}

int main()
{
    int save[2];
    int cnt=0;
    memset(save,-1,sizeof(save));
    scanf("%d %d %d",&M,&N,&step);
    for(int i=0;i<N;++i){
        for(int j=0;j<M;++j){
            scanf("%d",&canvas[i][j]);
            ++flag[canvas[i][j]];
        }
    }
    for(int i=0;i<N;++i){
        for(int j=0;j<M;++j){
            if(flag[canvas[i][j]]>1)continue;
            else if(examine(i,j)){
                if(++cnt>=2){
                    printf("Not Unique\n");
                    return 0;
                }
                save[0]=j;
                save[1]=i;
            }
        }
    }
    if(save[0]==-1)printf("Not Exist\n");
    else printf("(%d, %d): %d\n",save[0]+1,save[1]+1,canvas[save[1]][save[0]]);
    return 0;
}

1069

如果用一个数组取映射的话空间都炸了，所以改用map映射

#include <stdio.h>
#include <string>
#include <map>
#include <iostream>
using namespace std;
map<string,int> cal;

int main()
{
    int M,N,S;
    string bf;
    scanf("%d %d %d",&M,&N,&S);
    int cnt=S;
    for(int i=1;i<=M;++i){
        cin>>bf;
        if(i==cnt){
            if(cal[bf]==0){
                cout<<bf<<endl;
                cal[bf]=1;
                cnt+=N;
            }
            else cnt++;
        }
    }
    if(S>M)printf("Keep going...\n");
    return 0;
}

1070

基础的贪心法

#include <stdio.h>
#include <algorithm>
#include <iostream>
using namespace std;


int main()
{
    int N;
    int save[100000];
    scanf("%d",&N);
    for(int i=0;i<N;++i){
        scanf("%d",&save[i]);
    }
    sort(save,save+N);
    int sum=save[0];
    for(int i=1;i<N;++i){
        sum=(sum+save[i])/2;
    }
    printf("%d",sum);
    return 0;
}

1071

输出格式有错误。。。少了个空格

#include <iostream>
#include <stdio.h>
using namespace std;

int main()
{
    int T,K;
    int n1,b,n2;
    long t;
    cin>>T>>K;
    for(int i=0;i<K;++i){
        cin>>n1>>b>>t>>n2;
        if(t>T){
            printf("Not enough tokens.  Total = %ld.",T);
            continue;
        }
        if((b==0&&n1>n2)||(b==1&&n1<n2)){
            T+=t;
            printf("Win %ld!  Total = %ld.",t,T);
        }else{
            T-=t;
            printf("Lose %ld.  Total = %ld.",t,T);
            
        }
        if(T==0){
                printf("Game Over.");
                break;
        }
    }
    return 0;
}

1072

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
bool objects[10000];
char name[5];

int main()
{
    memset(objects,0,sizeof(objects));
    int N,M,temp,num=0,peo=0;
    scanf("%d %d",&N,&M);
    for(int i=0;i<M;++i){
        scanf("%d",&temp);
        objects[temp]=true;
    }
    int size,object;
    for(int j=0;j<N;++j){
        int save[10],cnt=0;
        scanf("%s %d",name,&size);
        for(int i=0;i<size;++i){
            scanf("%d",&object);
            if(objects[object]){
                save[cnt++]=object;
            }
        }
        if(cnt!=0){
            peo++;
            num+=cnt;
            printf("%s:",name);
            for(int i=0;i<cnt;++i)printf(" %04d",save[i]);
            putchar('\n');
        }
    }
    printf("%d %d\n",peo,num);
    return 0;
}

1073

令人害怕的大模拟题.jpg

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <string>
#define charToNum(x) (x-'a')
using namespace std;
struct prob{

    int score;
    int chioce;
    int isTrue;
    bool ans[5];
}probs[1000];

int cnt_wrong[1000][5];
int max_num=0;

float grade(bool answer[],int num){
    bool flag1=true;
    bool flag2=true;
    for(int i=0;i<probs[num].chioce;++i){
        if(probs[num].ans[i]==false&&answer[i]==true){
            max_num=max(max_num,++cnt_wrong[num][i]);
            flag1=false;
        }
        if(probs[num].ans[i]==true&&answer[i]==false){
            max_num=max(max_num,++cnt_wrong[num][i]);
            flag2=false;
        }
    }
    if(!flag1)return 0;
    if(flag2) return probs[num].score;
    return ((float)probs[num].score)/2;
}

int main()
{
    int N,M;
    char choose;
    scanf("%d %d",&N,&M);
    for(int i=0;i<M;++i){
        memset(probs[i].ans,0,sizeof(probs[i].ans));
        memset(cnt_wrong[i],0,sizeof(cnt_wrong[i]));
        cin>>probs[i].score>>probs[i].chioce>>probs[i].isTrue;
        for(int j=0;j<probs[i].isTrue;++j){
            cin>>choose;
            probs[i].ans[charToNum(choose)]=true;
        }
    }
    getchar();
    for(int i=0;i<N;++i){
        float grades=0;
        char c;
        bool save[5];
        memset(save,0,sizeof(save));
        int cnt=0;
        while((c=getchar())!='\n'){
            if(c>='a'&&c<='e')save[charToNum(c)]=true;
            else if(c==')'){
                grades+=grade(save,cnt++);
                memset(save,0,sizeof(save));
            }
        }
        printf("%.1f\n",grades);
    }
    if(max_num==0){
        printf("Too simple\n");
        return 0;
    }
    for(int i=0;i<M;++i){
        for(int j=0;j<probs[i].chioce;++j){
            if(max_num==cnt_wrong[i][j])
                printf("%d %d-%c\n",max_num,i+1,j+'a');
        }
    }
    return 0;
}

1074

reverse函数真的炒鸡好用呀

#include <iostream>
#include <string>
#include <algorithm>
#include <string.h>
#include <stdio.h>
#define charToNum(x) (x-'0')
using namespace std;
string a,b,index1;
int sum[20];
int pos=0,pos_a=0,pos_b=0;

void trans(int x){
    if(x==0)x=10;
    int temp=0;
    if(pos_a<a.size()&&pos_b<b.size())
        temp=charToNum(a[pos_a++]) + charToNum(b[pos_b++]);
    else if(pos_a==a.size())
        temp=charToNum(b[pos_b++]);
    else if(pos_b==b.size())
        temp=charToNum(a[pos_a++]);
    sum[pos]+=temp;
    sum[pos+1]+=sum[pos]/x;
    sum[pos++]%=x;
}

int main()
{
    memset(sum,0,sizeof(sum));
    cin>>index1>>a>>b;
    reverse(index1.begin(),index1.end());
    reverse(a.begin(),a.end());
    reverse(b.begin(),b.end());
    int max=a.size()>b.size()?a.size():b.size();
    for(int i=0;i<max;++i){
        trans(charToNum(index1[i]));
    }
    while(sum[pos]==0&&pos>0)pos--;
    for(;pos>=0;pos--){
        printf("%d",sum[pos]);
    }
    return 0;
}

1075

柳神的答案写的真好，用一个哈希映射表储存链表让时间复杂度大大降低，写出来的代码也简洁很多。

#include <iostream>
#include <stdio.h>
#include <vector>
using namespace std;
struct node{
    int next;
    int data;
}list[100000];

vector<int> v[3];

int main()
{
    int head,N,K;
    scanf("%d %d %d",&head,&N,&K);
    int pos;
    for(int i=0;i<N;++i){
        scanf("%d",&pos);
        scanf("%d%d",&list[pos].data,&list[pos].next);
    }
    while(head!=-1){
        if(list[head].data<0)
            v[0].push_back(head);
        else if(list[head].data<=K)
            v[1].push_back(head);
        else
            v[2].push_back(head);
        head=list[head].next;
    }
    bool flag=true;
    for(int i=0;i<3;++i){
        for(int j=0;j<v[i].size();++j){
            if(flag){
                printf("%05d %d",v[i][j],list[v[i][j]].data);
                flag=false;
            }
            else{
                printf(" %05d\n%05d %d",v[i][j],v[i][j],list[v[i][j]].data);
            }
        }
    }
    printf(" -1");
    return 0;
}