uva10976

这个，枚举到2n就可以了，其实还有一些小技巧，自己看吧

#include<cstdio>#include<algorithm>#include<cstring>#include<cmath>using namespace std;void cal(double k){    int n = 0;    for(double i = k + 1 ; i <= k * 2 ; i++)        if(k*i/(k-i) == floor(k*i/(k-i))){            n++;    }    printf("%d\n",n);    for(double i = k + 1 ; i <= k * 2 ; i++){        if(k*i/(k-i) == floor(k*i/(k-i))){            printf("1/%.0lf = 1/%.0lf + 1/%.0lf\n",k,k*i/(i-k),i);        }    }}int main(){    double k;    while(~scanf("%lf",&k)){        cal(k);    }    return 0;}