Uva10976

Fractions Again?! UVA - 10976

It is easy to see that for every fraction in the form 1 k (k > 0), we can always find two positive integers x and y, x ≥ y, such that: 1 k = 1 x + 1 y Now our question is: can you write a program that counts how many such pairs of x and y there are for any given k?
Input Input contains no more than 100 lines, each giving a value of k (0 < k ≤ 10000).
Output
For each k, output the number of corresponding (x,y) pairs, followed by a sorted list of the values of x and y, as shown in the sample output.
Sample Input
2 12
Sample Output
2 1/2 = 1/6 + 1/3 1/2 = 1/4 + 1/4 8 1/12 = 1/156 + 1/13 1/12 = 1/84 + 1/14 1/12 = 1/60 + 1/15 1/12 = 1/48 + 1/16 1/12 = 1/36 + 1/18 1/12 = 1/30 + 1/20 1/12 = 1/28 + 1/21 1/12 = 1/24 + 1/24

#include <cstdio>
#include <iostream>
#include <queue>
#include <vector>
#include<string.h>
#include<map>
#include<bits/stdc++.h>
#define LL long long
#define maxn 1005
using namespace std;
//vector<int> m;
int main()
{
    int k;
    while(scanf("%d",&k)!=EOF)
    //while(~scanf("%d",&k),k)这么写就WA，这是什么神奇的卡点，我表示很迷.jpg
    {
        int x[maxn],y[maxn];
        int ans=0;
        for(int i=k+1;i<=2*k;i++)//枚举y
        {
            int m=(k*i)/(i-k);
            if((k*i)%(i-k)==0&&m>=i)
                {
                    x[ans]=m,y[ans]=i;
                    ans++;
                }
        }
        printf("%d\n",ans);
        for(int i=0;i<ans;i++)
            printf("1/%d = 1/%d + 1/%d\n",k,x[i],y[i]);
        }
    return 0;
}

思路：

Cuz: x>=y 则1/x<=1/y 则1/k=1/x+1/y<=2/y

So: y<=2k && y>k,x>=y(由题目得)

x=(k*y)/(y-k),如果满足(k*y)%(y-k)则存在这样的正整数x。