PAT 甲级 1010 Radix

https://pintia.cn/problem-sets/994805342720868352/problems/994805507225665536

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

代码：

#include <bits/stdc++.h>
using namespace std;

string N1, N2;
int radix, tag;
long long sum = 0;

long long Pow(long long a, long long b) {
    long long ans1 = 1;

    while(b) {
        if(b % 2) {
            ans1 = ans1 * a;
            b --;
        } else {
            a = a * a;
            b /= 2;
        }
    }
    return ans1;
}

long long num(string s, int system) {
    int ls = s.length();
    reverse(s.begin(), s.end());
    long long ans = 0;
    if(system <= 10) {
        for(int i = 0; i < ls; i ++)
            ans += (s[i] - ‘0‘) * Pow(system, i);
    } else {
        int temp;
        for(int i = 0; i < ls; i ++) {
            if(s[i] >= ‘0‘ && s[i] <= ‘9‘)
                temp = s[i] - ‘0‘;
            else temp = s[i] - ‘a‘ + 10;

            ans += temp * Pow(system, i);
        }
    }
    return ans;
}

long long Find(string s, long long res) {
    char it = *max_element(s.begin(), s.end());
    long long l = (isdigit(it) ? it - ‘0‘: it - ‘a‘ + 10) + 1;
    long long r = max(res, l);
    long long mid;

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

int main() {
    cin >> N1 >> N2 >> tag >> radix;
    int l1 = N1.length(), l2 = N2.length();
    long long out = 0;
    if(tag == 1) {
        sum = num(N1, radix);
        out = Find(N2, sum);
    } else {
        sum = num(N2, radix);
        out = Find(N1, sum);
    } 

    if(out == -1) printf("Impossible\n");
    else printf("%lld\n", out);
    return 0;
}

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原文地址：https://www.cnblogs.com/zlrrrr/p/10350969.html