题目描述

Given a pair of positive integers, for example, 6 and 110, can this equation 6 = 110 be true? The answer is yes, if 6 is a decimal number and 110 is a binary number.

Now for any pair of positive integers N​1​​ and N​2​​, your task is to find the radix of one number while that of the other is given.

输入描述

Each input file contains one test case. Each case occupies a line which contains 4 positive integers:
N1 N2 tag radix

Here N1 and N2 each has no more than 10 digits. A digit is less than its radix and is chosen from the set { 0-9, a-z } where 0-9 represent the decimal numbers 0-9, and a-z represent the decimal numbers 10-35. The last number radix is the radix of N1 if tag is 1, or of N2 if tag is 2.

输出描述

For each test case, print in one line the radix of the other number so that the equation N1 = N2 is true. If the equation is impossible, print Impossible. If the solution is not unique, output the smallest possible radix.

输入例子1

6 110 1 10

输出例子1

2

输入例子2

1 ab 1 2

输出例子2

Impossible

我的代码

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm> 
using namespace std;
typedef long long LL;
long long Map[256]; 
long long inf=(1LL << 63)-1;

void init(){  //对'0'-'9' 'a'-'z'进行初始化赋值 
    for(char c='0';c<='9';c++){
        Map[c]=c-'0';
    } 
    for(char c='a';c<='z';c++){
        Map[c]=c-'a'+10;   
    }
}

long long convertNum10(char a[],long long radix,long long t){  //转换为10进制 
    long long ans=0;
    int len=strlen(a);
    for(int i=0;i<len;i++){
        ans=ans*radix+Map[a[i]];
        if(ans<0 || ans>t) return -1;  //判断溢出 
    } 
    return ans;
}

long long findLargestDigit(char n2[]){  //找到n2的下界 
    int ans=-1,len=strlen(n2);
    for(int i=0;i<len;i++){
        if(Map[n2[i]]>ans){
            ans=Map[n2[i]];
        }
    } 
    return ans+1;
}

int cmp(char n2[],long long radix,long long t){  //将n2转换为10进制之后与t进行判断 
    int len=strlen(n2);
    long long num=convertNum10(n2,radix,t);
    if(num<0) return 1; 
    if(t>num) return -1;
    else if(t==num) return 0;
    else return 1;
}

long long binarySearch(char n2[],long long left,long long right,long long t){  //二分法查找进制 
    long long mid;
    while(left<=right){
        mid=(left+right)/2;
        int flag=cmp(n2,mid,t);
        if(flag==0) return mid;
        else if(flag==-1) left=mid+1;
        else right=mid-1;
    } 
    return -1;
}


int main(){
    char n1[20],n2[20],temp[20];
    int tag,radix;
    cin>>n1;
    cin>>n2;
    cin>>tag;
    cin>>radix;
    init();  
    if(tag==2){  //将确定进制的数放在n1 
        strcpy(temp,n1);
        strcpy(n1,n2);
        strcpy(n2,temp);
    }
    long long t=convertNum10(n1,radix,inf);  //将n1转换为10进制数 
    long long low=findLargestDigit(n2);  //找到n2中最大的数 
    long long high=max(t,low)+1;  
    long long ans=binarySearch(n2,low,high,t); //求解n2的进制 
    if(ans==-1) cout<<"Impossible";
    else cout<<ans;
    return 0;
}