UVa 10976 Fractions Again?

题目链接：

http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=37234

题目描述：

输入正整数k，找到所有正整数 x>=y,使得1/k = 1/x + 1/y.

思路：

因为 x>=y 所以有: 1/k <= 1/y + 1/y –> 1/k <= 2/y –> y <= 2k. 又因为 1/k = 1/y + 1/x –> k*y / (y-k) = x,　即当　k*y % (y-k) == 0时，必然得整数x，此对xy为一解。

/*************************************************************************    > File Name: UVa_10976.cpp    > Author: dulun    > Mail: [e-mail]    > Created Time: 2016年03月19日 星期六 00时45分43秒 ************************************************************************/#include<iostream>#include<stdio.h>#include<cstring>#include<cstdlib>#include<algorithm>#define LL long longusing namespace std;const int N = 1086;int x[N];int Y[N];int main(){    int k;    while(~scanf("%d", &k))    {        int sum = 0;        int k2 = k<<1;        for(int y = k+1; y <= k2; y++)        {            if(k * y % (y - k) == 0)            {                x[sum] = k*y / (y-k);                Y[sum] = y;                sum++;            }        }        printf("%d\n", sum);        for(int i = 0; i < sum; i++) printf("1/%d = 1/%d + 1/%d\n", k, x[i], Y[i]);    }    return 0;}