【CUGBACM15级BC第9场 A】hdu 4993 Revenge of ex-Euclid

Revenge of ex-Euclid

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 877    Accepted Submission(s): 531

Problem Description

In arithmetic and computer programming, the extended Euclidean algorithm is an extension to the Euclidean algorithm, which computes, besides the greatest common divisor of integers a and b, the coefficients of Bézout's identity, that is integers x and y such that ax + by = gcd(a, b).
---Wikipedia

Today, ex-Euclid takes revenge on you. You need to calculate how many distinct positive pairs of (x, y) such as ax + by = c for given a, b and c.

 

Input

The first line contains a single integer T, indicating the number of test cases.

Each test case only contains three integers a, b and c.

[Technical Specification]
1. 1 <= T <= 100
2. 1 <= a, b, c <= 1 000 000

 

Output

For each test case, output the number of valid pairs.

 

Sample Input

21 2 31 1 4

 

Sample Output

13

 

题意：
寻找ax+by = c有多少组解！

#include<bits/stdc++.h>#include <ctime>using namespace std;typedef long long ll;const int MAXN = 1 * 1e5 + 500;const ll M = 1e9 + 7;int main(){    ///clock_t start_time = clock();    ///clock_t end_time = clock();    ///cout << "Running time is: " << static_cast<double>(end_time - start_time) / CLOCKS_PER_SEC * 1000 << "ms" << endl;    std::ios::sync_with_stdio(false);    int t;    cin >> t;    while (t--)    {        int a, b, c, cnt = 0;        cin >> a >> b >> c;        for (int x = 1; x * a < c; x++)        {            if ((c - a*x) % b == 0)            {                cnt++;            }        }        cout << cnt << endl;    }    return 0;}