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
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
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9
return 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;
}