HDU 4993 revenge of ex-Euclid)

Question link: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4993

Problem descriptionin 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 bé Zout'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.
Inputthe 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
Outputfor each test case, output the number of valid pairs.
Sample Input

21 2 31 1 4

 
Sample output

13

 
Sourcebestcoder round #9

Question:

How many group solutions does AX + by = C have!

The Code is as follows:

#include <cstdio>#include <cmath>int main(){    int t;    int a,b,c;    scanf("%d",&t);    while(t--)    {        scanf("%d%d%d",&a,&b,&c);        int k=0;        for(int x = 1; x*a < c; x++)        {            if((c-a*x)%b == 0)                k++;        }        printf("%d\n",k);    }    return 0;}

HDU 4993 revenge of ex-Euclid)