（UVA - 10976）Fractions Again?!（技巧，暴力枚举）

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

题意：给定正整数k

（0<k<=10000）

，找到所有的正整数x>=y 使1/k=1/x+1/y输出等式

分析：x>=y -> 1/x<=1/y

又1/x=1/k-1/y -> 1/k-1/y<=1/y -> 1/k<=2/y -> y<=2k

等式中隐含条件 y>=k

故枚举 k<=y<=2k，当然x不能枚举，直接计算出来是否满足等式，枚举的话TLE

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1e3+5;
int a
,b
;
int main()
{
int k;
while(~scanf("%d",&k))
{
int y,cnt=0;
for(y=k;y<=2*k;y++)
{
double x=1.0*k*y/(y-k);
if(x==(int)x)
{
a[cnt]=(int)x;
b[cnt]=y;
cnt++;
}
}
printf("%d\n",cnt);
for(int i=0;i<cnt;i++)
printf("1/%d = 1/%d + 1/%d\n",k,a[i],b[i]);
}
return 0;
}