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

题目分析：将两个元素转换为10进制进行比较 ，对于那个不确定进制的参数，我们可以使用二分法来求它的进制
参考了柳神的答案

#include<iostream>
#include<string>
#include<stdlib.h>
#include<vector>
#include<algorithm>
using namespace std;

long long tolong(string str, long long radix)
{
    long long length = str.length();
    long long num = 0;
    for (long long j = 0; j < length; j++)
        num = (isdigit(str[j])) ? num * radix + str[j] - '0' : num * radix + str[j] - 'a' + 10;
    return num;
}
long long Find_radix(string num,long long N)
{
    char it = *max_element(num.begin(), num.end());
    long long low = ((isdigit(it)) ? it - '0' : it - 'a' + 10)+1;
    long long high = max(N,low);
    long long mid;
    while (low<=high)
    {
        mid = (low + high) / 2;
        long long n = tolong(num, mid);
        if (n > N||n<0)
            high = mid-1;
        else if (n < N)
            low = mid+1;
        else
            return mid;
    }
    return -1;
}

int main()
{
    string N1, N2;
    long long tag, radix;
    cin >> N1 >> N2 >> tag >> radix;
    long long flag = 0;
    flag = tag == 1 ? Find_radix(N2, tolong(N1, radix)) : Find_radix(N1, tolong(N2, radix));
    if (flag==-1)
        cout << "Impossible";
    else
        cout << flag;
    return 0;
}