A1010 Radix (25分)

最新推荐文章于 2022-03-23 18:17:03 发布

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最新推荐文章于 2022-03-23 18:17:03 发布

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本文探讨了如何解决两个正整数在不同基数下相等的问题，通过基数转换算法，实现了从已知数的基数到未知数基数的求解。输入包含两个正整数和其中一个数的基数，输出为使等式成立的另一个数的最小可能基数，或在不可能的情况下输出'Impossible'。

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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

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;
}