SDNU——8题——B

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<cstdio>#include<iostream>#include<vector>using namespace std;int main(){vector<int> rx;vector<int> ry;int k,x,y,cnt=0,l=0,i;while(cin>>k){for(y=k+1;y<=2*k;y++){if((k*y)%(y-k)==0&&(k*y)/(y-k)>=y){x=(k*y)/(y-k);cnt++;rx.push_back(x);ry.push_back(y);l++;}}cout<<cnt<<"\n";for(i=0;i<l;i++){printf("1/%d = 1/%d + 1/%d\n",k,rx[i],ry[i]);}cnt=0;rx.clear();ry.clear();l=0;}return 0;}

1.暴力要有度，所以想要暴力x与y就得找到，他们的限制之处，这样暴力才能有个头，才不会超时~

你可以试一试你会求出y的范围是【k,2k】，找到y的范围后，我没能找到x的具体范围，所以我就把所有x的值找出来，1.判断是否能整除2.判断是否满足x>y

2.注意输出的一般格式

3.注意下一次输入时，把数组什么的都清空，防止对下一侧的计算造成误差~~~