## Describe

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``

## std

``````#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

ll res = 0;
for(int i = 0 ; i < str.length(); ++i){
if(str[i]=='-') continue;
if(str[i] >= '0' && str[i] <= '9')
res = res * radix + str[i] - '0';
else
res = res * radix + str[i]-'a' + 10;
}
if(str == '-') res *= -1;
return res;
}
int binary_find(string str,ll num){
char it = *max_element(str.begin(), str.end());
ll low = (isdigit(it) ? it - '0' : it - 'a' + 10) + 1;
ll high = max(num, low);
while(low <= high){
ll mid = (low + high) >> 1;
ll temp = to_dec(str, mid);
if(temp < 0 || temp > num) high = mid - 1;
else if(temp == num) return mid;
else low = mid + 1;
}
return -1;
}
int main(){
string n1, n2;
cin >> n1 >> n2 >> tag >> radix;
if(tag == 2) swap(n1,n2);
if(res_tag == -1) cout << "Impossible";
else cout << res_tag;
return 0;
}``````

Last modification：November 22nd, 2019 at 08:02 pm 