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[];

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

 void init(){

     for(char c='';c<='';c++)
         digit[c] = c-'';

     for(char c='a';c<='z';c++)
         digit[c] = c-'a'+;
 }

 long long convertNum10(string str1,long long radix,long long t){
     long long ans=;

     long long len = str1.size();

     for(long long i=;i<len;i++){
         char c=str1[i];
         ans=ans*radix+digit[c];

         if(ans<||ans>t) return -;
     }

     return ans;

 }

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

 }

 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)/;

         long long flag=cmp(str2,mid,t);

         if(flag<)left =mid+;
         else if(flag>)right = mid-;
         else return mid;
     }

     return -;

 }

 long long findLargest(string &str){
     long long ans=-;
     long long len=str.size();

     for(long long i=;i<len;i++){
         if(digit[str[i]]>ans)
             ans=digit[str[i]];
     }

     return ans+;

 }

 int main(){
     string str1,str2;
     long long tag,radix;

     cin>>str1>>str2>>tag>>radix;

     init();

     if(tag==)
         swap(str1,str2);

     long long t=convertNum10(str1,radix,inf);

     long long low = findLargest(str2);

     long long high=max(low,t)+; 

     long long ans=binary_search(str2,low,high,t);

     if(ans==-)cout<<"Impossible\n";
     else cout<<ans<<endl;

 }