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 N1 and N2, 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:

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;

 }

巴特西