HDU 4993 Revenge of ex-Euclid（BestCoder Round #9）

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

题意：给出a，b，c，输出满足ax+by==c的（x，y）的个数。

利用扩展欧几里得找出该线性方程的其中一个解（x0，y0），然后可以推广得到其他的解：

a*x0+b*y0=c；

a*（x0+b）+b*（y0-a）=c；

a*（x0+2*b）+b*（y0-2*a）=c；

......

因为该题要求所有正整数解的个数，所以需要通过我们求出的这一个解，寻找边界条件，进而得到解的个数。

#include<stdio.h>
#include<string.h>
#include<queue>
#include<math.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;

const int N=1e6+10;
const int INF=0x3f3f3f3f;
const int MOD=1e9+7;

typedef long long LL;

int x, y;

int Ex_gcd(int a, int b)
{
    int t, z;

    if (b == 0)
    {
        x = 1;
        y = 0;

        return a;
    }

    z = Ex_gcd(b, a%b);

    t = x;
    x = y;
    y = t-(a/b)*y;

    return z;
}

int main ()
{
    int T, a, b, c, ans, z;

    scanf("%d", &T);

    while (T--)
    {
        scanf("%d%d%d", &a, &b, &c);

        ans = 0;

        z = Ex_gcd(a, b); ///得到公约数

        a /= z; b /= z; c /= z;

        z = Ex_gcd(a, b);

        x *= c; y *= c; ///找到满足方程的最小解

        while (x - b > 0) ///找到最小的>0的x
        {
            x -= b;
            y += a;
        }
        ///那么此时y肯定>0
        while (y > 0) ///每次让y-a，a+b，那么此时x肯定满足条件，当y<=0时就不满足条件了
        {
            if (x > 0) ans++; ///x可能在一开始是负数

            x += b;
            y -= a;
        }

        printf("%d\n", ans);
    }

    return 0;
}