uva 10976 Fractions Again(简单枚举)

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10976 Fractions Again

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>
#include <cstdio>
using namespace std;
int xx[1005],yy[1005];
int main()
{
    int k,x,y,total;
    
    while(cin >> k && k!=0)
    {
        total = 0;
        for(int i=k+1; i<=2*k; i++)
        {
            if(k*i%(i-k)==0)
            {
                xx[total] = k*i/(i-k);
                yy[total] = i;
                total++;
            }
        }
        cout<<total<<endl;
        for(int i=0; i<total; i++)
        {
            printf("1/%d = 1/%d + 1/%d\n",k,xx[i],yy[i]);
        }
    }
    return 0;
}

uva 10976 Fractions Again(简单枚举)

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原文地址：http://www.cnblogs.com/hfc-xx/p/4686460.html