1010 Radix (25 分)

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.

Input Specification:

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.

Output Specification:

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.

Sample Input 1:

6 110 1 10

Sample Output 1:

2

Sample Input 2:

1 ab 1 2

Sample Output 2:

Impossible

注意点：1.left<=right;
　　　　2.数据范围用long long

#include<bits/stdc++.h>
using namespace std;

long long digit[256];

const long long inf = (1LL<<63)-1;



void init(){
    
    for(char c='0';c<='9';c++)
        digit[c] = c-'0';
        
        
    for(char c='a';c<='z';c++)
        digit[c] = c-'a'+10;
}

long long convertNum10(string str1,long long radix,long long t){
    long long ans=0;
    
    long long len = str1.size();
    
    for(long long i=0;i<len;i++){
        char c=str1[i]; 
        ans=ans*radix+digit[c];
        
        if(ans<0||ans>t) return -1;        
    }
    
    return ans;
    
}


long long cmp(string str2,long long radix,long long t){
    long long n2=convertNum10(str2,radix,t);
    if(n2<0)return 1;
    else if(n2==t) return 0;
    else if(n2>t) return 1;
    else return -1;
    
}

long long binary_search(string &str2,long long low,long long high,long long t){
    long long left=low,right=high;
    long long mid;
    
    while(left<=right){
        mid=(left+right)/2;
        
        long long flag=cmp(str2,mid,t);
        
        if(flag<0)left =mid+1;
        else if(flag>0)right = mid-1;
        else return mid;
    }
    
    
    return -1;

}


long long findLargest(string &str){
    long long ans=-1;
    long long len=str.size();
    
    for(long long i=0;i<len;i++){
        if(digit[str[i]]>ans)
            ans=digit[str[i]];
    }
    
    
    return ans+1;
    
    
}


int main(){
    string str1,str2;
    long long tag,radix;
    
    cin>>str1>>str2>>tag>>radix;
    
    init();
    
    if(tag==2)
        swap(str1,str2);
        
        
    long long t=convertNum10(str1,radix,inf);
    
    long long low = findLargest(str2);
    
    long long high=max(low,t)+1; 
    
    long long ans=binary_search(str2,low,high,t);
    
    
    if(ans==-1)cout<<"Impossible
";
    else cout<<ans<<endl;
    
}