PAT 甲级 1010 Radix（25 分）

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本文介绍了一个算法问题——判断两个不同进制的数是否相等，并提供了完整的代码实现。输入包括两个正整数及其对应的进制，输出则是另一个数的进制，使得两数相等。若无解，则输出 Impossible。

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

#include <iostream>
#include <cctype>
#include <algorithm>
#include <cmath>
using namespace std;
long long convert(string n, long long radix) {
    long long sum = 0;
    int index = 0, temp = 0;
    for (auto it = n.rbegin(); it != n.rend(); it++) {
        temp = isdigit(*it) ? *it - '0' : *it - 'a' + 10;
        sum += temp * pow(radix, index++);
    }
    return sum;
}
long long find_radix(string n, long long num) {
    char it = *max_element(n.begin(), n.end());
    long long low = (isdigit(it) ? it - '0': it - 'a' + 10) + 1;
    long long high = max(num, low);
    while (low <= high) {
        long long mid = (low + high) / 2;
        long long t = convert(n, mid);
        if (t < 0 || t > num) high = mid - 1;
        else if (t == num) return mid;
        else low = mid + 1;
    }
    return -1;
}
int main() {
//	freopen("C:\\1.txt", "r", stdin);
    string n1, n2;
    long long tag = 0, radix = 0, result_radix;
    cin >> n1 >> n2 >> tag >> radix;
    result_radix = tag == 1 ? find_radix(n2, convert(n1, radix)) : find_radix(n1, convert(n2, radix));
    if (result_radix != -1) {
        printf("%lld", result_radix);
    } else {
        printf("Impossible");
    }   
    return 0;
}