B1082 Read Number in Chinese (25分)

• Given an integer with no more than 9 digits, you are supposed to read it in the traditional Chinese way. Output Fu first if it is negative.
  For example, -123456789 is read as Fu yi Yi er Qian san Bai si Shi wu Wan liu Qian qi Bai ba Shi jiu. Note: zero (ling) must be handled correctly according to the Chinese tradition. For example, 100800 is yi Shi Wan ling ba Bai.

我的思路：

所有输入的字符串往前补0补到九位，按9位处理
按各部分全零非全零分，有2 * 2 * 2种，然后读上“万”，“千百十个”整体拿出来处理，
//亿位有非0数，万位不足四位时前面不能读0
//“千百十个”不会处理，我就把所有情况罗列出来了。。。🤦‍“万”的个位可以读0，后面的部分个位不能读0，每个部分的所有前导0都只读一次。这样的思路还是枚举法，很容易少考虑到一些情况，如果实战真用这种方法，那心态估计早就崩了吧。。。。。。

网上的思路：

学习的代码：

#include <iostream>
#include <string>
#include <vector>
using namespace std;
string num[10] = { "ling","yi", "er", "san", "si", "wu", "liu", "qi", "ba", "jiu" };
string c[6] = { "Ge","Shi", "Bai", "Qian", "Yi", "Wan" };
int J[] = {1, 10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000};
vector<string> res;
int main() {
    int n;
    cin >> n;
    if (n == 0) {
        cout << "ling";
        return 0;
    }
    if (n < 0) {
        cout << "Fu ";
        n = -n;
    }
    int part[3];
    part[0]= n / 100000000; 
    part[1]= (n % 100000000) / 10000;
    part[2] = n % 10000;
    bool zero = false; //是否在非零数字前输出合适的ling
    int printCnt = 0; //用于维护单词前没有空格，之后输入的单词都在前面加一个空格。
    for (int i = 0; i < 3; i++) {
        int temp = part[i]; //三个部分，每部分内部的命名规则都一样，都是X千X百X十X
        for (int j = 3; j >= 0; j--) {
            int curPos = 8 - i * 4 + j; //当前数字的位置
            if (curPos >= 9) continue; //最多九位数
            int cur = (temp / J[j]) % 10;//取出当前数字
            if (cur != 0) {
                if (zero) {
                    printCnt++ == 0 ? cout<<"ling" : cout<<" ling";
                    zero = false;
                }
                if (j == 0)
                    printCnt++ == 0 ? cout << num[cur] : cout << ' ' << num[cur]; //在个位，直接输出
                else                             
                    printCnt++ == 0 ? cout << num[cur] << ' ' << c[j] : cout << ' ' << num[cur] << ' ' << c[j]; //在其他位，还要输出十百千
            } else {
                if (!zero && j != 0 && n / J[curPos] >= 10) zero = true;   //注意100020这样的情况
            }
        }
        if (i != 2 && part[i]>0) cout << ' ' << c[i + 4]; //处理完每部分之后，最后输出单位，Yi/Wan
    }
    return 0;
}

“千百十个”需要改进优化！

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string.h>
#include <cmath>
#include <math.h>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <stack>
using namespace std;
typedef long long ll;
const int MAX=1005;
const int INF=0x3f3f3f3f;
const int mod=1000000007;
char mp[8][10]={"Yi","Qian","Bai","Shi"};
char num[10][10]={"ling","yi","er","san","si","wu","liu","qi","ba","jiu"};
void du(string s,int flag)
{
    int i=0;
    while(s[i]=='0')//已剔除所有0的情况
    {
        i++;
    }
    if(i!=0)
    {
        if(flag==0)
           cout<<"ling"<<" ";
        if(i==1)
        {
            cout<<num[s[1]-'0']<<" "<<mp[2];
              if(s[2]=='0'&&s[3]!=0)
            {
                cout<<" "<<"ling"<<" "<<num[s[3]-'0'];
            }
            else if(s[2]!='0'&&s[3]=='0')
            {
                cout<<" "<<num[s[2]-'0']<<" "<<mp[3];
            }
            else if(s[2]!='0'&&s[3]!='0')
            {
                cout<<" "<<num[s[2]-'0']<<" "<<mp[3]<<" "<<num[s[3]-'0'];
            }
        }
        else if(i==2)
        {
            if(s[3]=='0')
            cout<<num[s[2]-'0']<<" "<<mp[3];
            else
            cout<<num[s[2]-'0']<<" "<<mp[3]<<" "<<num[s[3]-'0'];
        }
        else if(i==3)
             cout<<num[s[3]-'0'];
    }
    else
    {
        if(s[3]!='0')
        {
            if(s[1]=='0'&&s[2]=='0')//1001型
            {
                cout<<num[s[0]-'0']<<" "<<mp[1]<<" "<<"ling";
            }
            else//1011型
            {
                for(int j=0;j<3;j++)
                {
                    if(j!=0)
                    {
                        cout<<" ";
                    }
                    if(s[j]=='0')
                        cout<<"ling";
                    else
                        cout<<num[s[j]-'0']<<" "<<mp[j+1];
                }
            }
             cout<<" "<<num[s[3]-'0'];
        }
        else if(s[3]=='0')
        {
            cout<<num[s[0]-'0']<<" "<<mp[1];
            if(s[1]=='0'&&s[2]!=0)
            {
                cout<<" "<<"ling"<<" "<<num[s[2]-'0']<<" "<<mp[3];
            }
            else if(s[1]!='0'&&s[2]!='0')
            {
                cout<<" "<<num[s[1]-'0']<<" "<<mp[2]<<" "<<num[s[2]-'0']<<" "<<mp[3];
            }
            else if(s[1]!='0'&&s[2]=='0')
            {
                cout<<" "<<num[s[1]-'0']<<" "<<mp[2];
            }
        }
    }
}
int main()
{
    string s;
    cin>>s;
    if(s[0]=='-')
    {
        cout<<"Fu ";
        s.erase(s.begin());
    }
    while(s.size()<9)
    {
        s='0'+s;
    }
    if(s=="000000000")
        cout<<"ling";
    else
    {
    int flagwan=0,flagqian=0;
    int check=0;
    if(s[0]!='0')
    {
        cout<<num[s[0]-'0']<<" "<<"Yi";
        check=1;
    }
    for(int i=1;i<=4;i++)
    {
        if(s[i]!='0')
            flagwan=1;//万位有数字
    }
    for(int i=5;i<=8;i++)
    {
            if(s[i]!='0')
                flagqian=1;//千位有数字
    }
        if(flagwan==1&&flagqian==1)
        {
            if(check==1)
            {
                cout<<" ";
                du(s.substr(1,4),0);
            }
            else
                du(s.substr(1,4),1);
            cout<<" Wan ";
            du(s.substr(5,8),0);
        }
        else if(flagwan==1&&flagqian==0)
        {
             if(check==1)
             {
                cout<<" ";
                du(s.substr(1,4),0);
             }
            else
                du(s.substr(1,4),1);
             cout<<" Wan ";
        }
        else if(flagwan==0&&flagqian==1)
             du(s.substr(5,8),0);
    }
  return 0;
}