Fractions Again?!（简单枚举）

最新推荐文章于 2020-03-10 16:22:37 发布

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最新推荐文章于 2020-03-10 16:22:37 发布

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本文介绍了一个有趣的数学问题：如何编程找出特定形式分数1/k的所有可能的正整数对(x, y)，使得1/k = 1/x + 1/y。文章提供了一个有效的算法，并通过示例输入输出展示了该算法的应用。

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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:

.
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

/// 1/n = 1/x + 1/y ==> n*y + n*x = x*y;
///y只需要从n+1枚举到2*n即可，此时x = n*y / (y-n);
#include<stdio.h>
#include<string.h>
int main()
{
    int n,i,p,num;
    int a[10000],b[10000];
    while(~scanf("%d",&n))
    {
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        num=0;
        for(i=n+1; i<=n*2; i++)
        {
            if(n*i%(i-n))continue;
            p=n*i/(i-n);
            a[num]=p;
            b[num]=i;
            num++;
        }
        printf("%d\n",num);
        for(i=0; i<num; i++)
            printf("1/%d = 1/%d + 1/%d\n",n,a[i],b[i]);
    }
}