A1010 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

Solution: 

#include <iostream>
#include <map>
#include <algorithm>

using namespace std;

map<char, int> mp;

void init() {
    for (char c = '0'; c <= '9'; c++) {
        mp[c] = c-'0';
    }
    for (char c = 'a'; c <= 'z'; c++) {
        mp[c] = c-'a'+10;
    }
}

long long convert(string know, long long radix) {
    long long num = 0;
    for (int i = 0; i < know.length(); i++) {
        num = num*radix + mp[know[i]];
    }
    return num;
}

long long find_radix(string unknow, long long num) {
    char min_ch = *max_element(unknow.begin(), unknow.end());
    long long l = mp[min_ch] + 1;
    long long r = max(num, l);

    while (l <= r) {
        long long mid = (l+r)/2;
        long long unknow_val = convert(unknow, mid);
        if (unknow_val == num) {
            return mid;
        } else if (unknow_val < 0 || unknow_val > num) {
            r = mid - 1;
        } else {
            l = mid + 1;
        }
    }
    return -1;
}

int main() {
    init();
    string A, B;
    long long tag, radix, result;
    cin >> A >> B >> tag >> radix;
    if (tag == 1) {
        result = find_radix(B, convert(A, radix));
    } else {
        result = find_radix(A, convert(B, radix));
    }
    if (result == -1) {
        cout << "Impossible";
    } else {
        cout << result;
    }
    return 0;
}