Fractions Again?! UVA - 10976

标签：show   inpu   ios   span   lin   div   for   size   string   

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 <iostream>//求1/n = 1/x + 1/y
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
using namespace std;
int x[10011],y[10011];
int main()
{
    int n,cnt;
    while(cin >> n)
    {
        memset(x,0,sizeof(x));
        memset(y,0,sizeof(y));
        cnt = 0;
        for(int i=n+1;i<=2*n;i++)
        {
            if((i*n)%(i-n)==0)//注意这个判断条件，判断等式是否可能成立
            {
                x[cnt] = (i*n)/(i-n);//求x的式子
                y[cnt] = i;
                cnt++;
            }
        }
        printf("%d\n",cnt);
        for(int i=0;i<cnt;i++)
            printf("1/%d = 1/%d + 1/%d\n",n,x[i],y[i]);
    }
    return 0;
}

Fractions Again?! UVA - 10976

标签：show   inpu   ios   span   lin   div   for   size   string   

原文地址：http://www.cnblogs.com/l609929321/p/6900911.html