HDU4993 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): 212 Accepted Submission(s): 120

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
2
1 2 3
1 1 4


Sample Output
1
3


Source
BestCoder Round #9

题意：输入a,b,c求a*x+b*y=c，有几组解。。。

第一眼看到这题就知道扩展欧几里得可以做。。。然后就用这个写了。。。。赛后看了下别人的。。。居然暴力就可以。。。

说下扩展欧几里得做法，扩展欧几里得求出一组解（a*x+b*y=gcd(a,b)），gcd(a,b)=d然后判断c能否整除d，不能无解。。有的话x+k*b/d,y-k*a/d（反过来也行），统计下都是正数的个数就行。。

#include
#include
#include
using namespace std;
void exgcd(int a,int b,int &d,int &x,int &y)
{
    if(!b)
        d=a,x=1,y=0;
    else
        {
        exgcd(b,a%b,d,y,x);
        y-=a/b*x;
        }
}
int main()
{
    int t,a,b,c;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d",&a,&b,&c);
        int d,x,y;
        int ans=0;
        exgcd(a,b,d,x,y);
        if(c%d)
        {
            printf("0\n");
            continue;
        }
        int k=c/d;
        x=x*k;
        y=y*k;
        int temp;
        int x1=x,y1=y;
        while(x1-b/d>0)
        {
            x1-=b/d;
            y1+=a/d;
        }
        while(y1>0)
        {
            if(x1>0)

                ans++;
            x1+=b/d;
            y1-=a/d;
        }
        printf("%d\n",ans);
    }
    return 0;
}