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; }