分数拆分（ Fractions Again， UVA 10976）-ACM

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 givenk?

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

212

Sample Output

21/2 = 1/6 + 1/31/2 = 1/4 + 1/481/12 = 1/156 + 1/131/12 = 1/84 + 1/141/12 = 1/60 + 1/151/12 = 1/48 + 1/161/12 = 1/36 + 1/181/12 = 1/30 + 1/201/12 = 1/28 + 1/211/12 = 1/24 + 1/24

这个题目做起来不难，难点在数值精度到问题上，我是参照了这为朋友到讲解

http://www.2cto.com/kf/201111/111420.html

/* * FractionAgain.cpp * *  Created on: 2014-8-27 *      Author: root */#include <iostream>#include <vector>#include <string>#include <cstdio>using namespace std;bool isInt(double n){double c = n-(int)n;if(n >= 0){if( c < 1e-15 || c < -0.999999999999999 ) {//单精度对应1e-6和6个9，双精度对应1e-15和15个9return true;}else{return false;}}else{ if( -c < 1e-15 || -c < -0.999999999999999 ){ return true; } else{ return false; }}}int main(){long  k ;vector<string> ans;char str[100];while((cin>>k)  && k != 0){long  max = k << 1;int y;ans.clear();for ( y = k + 1; y <= max; ++y) {double  x = (double)(k*y)/(y - k);if(isInt(x)){sprintf(str,"1/%ld = 1/%d + 1/%d\n",k,(int)x,y);ans.push_back(str);}}int size = ans.size();cout<<size<<endl;for (y = 0;y < size;y++) {cout<<ans[y];}}return 0;}