1010 Radix （25 分）

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

题意：如果tag=1，则N1是radix进制数，若tag=2，则N2是radix数，问是否存在某进制数使得N2（N1）为N1（N2）的进制数。

分析：因为进制数越大所得到的值越大，比如110，若是二进制，值为6，若为10进制，值为110。

依据这点二分：

1、首先看N1、N2谁大，若N2大交换下N1、N2顺序。2、二分下界为所出现数字中最大数字+1，上界为所给的radix，夹出对应进制数

 /**
 * Copyright(c)
 * All rights reserved.
 * Author : Mered1th
 * Date : 2019-02-26-14.28.46
 * Description : A1010
 */
 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<cmath>
 #include<algorithm>
 #include<string>
 #include<unordered_set>
 #include<map>
 #include<vector>
 #include<set>
 using namespace std;

 long long convert(string a,long long radix){ //转化为十进制
     ;
     ,len=a.size();
     ;i<len;i++){
         temp=isdigit(a[i])? a[i]-;
         sum=sum*radix+temp;
     }
     return sum;
 }

 long long find_radix(string a,long long num){
     //二分法，进制数越大数值越大。先确定上下界：下界是最大值+1，比如987，最起码是一个10进制数。而上界是max(下界，N1的十进制）+1；
     char it=*max_element(a.begin(),a.begin());
     )+;
     long long high=max(low,num);
     while(low<=high){
         ;
         long long t=convert(a,mid);
         ||t>num) high=mid-;
         else if(t==num) return mid;
         ;
     }
     ;
 }

 int main(){
 #ifdef ONLINE_JUDGE
 #else
     freopen("1.txt", "r", stdin);
 #endif
     string N1,N2;
     int tag;
     long long radix,ans;
     cin>>N1>>N2>>tag>>radix;
     ){
         swap(N1,N2);
     }
     ans=find_radix(N2,convert(N1,radix));
     ) {
         printf("%lld",ans);
     }else{
         printf("Impossible");
     }

     ;
 }