Fractions Again?!（暴力）

It is easy to see that for every fraction in the form  (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

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Problemsetter: Mak Yan Kei

AC代碼

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>

#define MAXN 10000
typedef long long ll;
ll a[MAXN], b[MAXN];

int main()
{
    ll k;
    while(scanf("%lld", &k) != EOF)
    {
        ll y = 1, cnt = 0;
        memset(a, 0, sizeof(a));
        memset(b, 0, sizeof(b));
        for(y = k + 1; y <= 2 * k; y++)
        {
            double x = 1.0 / (1.0 / k - 1.0 / y);
            ll xint = (ll)(x + 0.5);
            if(x >= 0 && fabs(x - xint) < 1e-4)
            {
                a[cnt] = xint;
                b[cnt] = y;
                cnt++;
            }
        }
        printf("%d\n", cnt);
        for(int i = 0; i < cnt; i++)
        {
            printf("1/%lld = 1/%lld + 1/%lld\n", k, a[i], b[i]);
        }
    }
    return 0;
}