369 combinations.cpp

来自「UVA solutions to few selected problems」· C++ 代码 · 共 26 行

#include<iostream>
#include<cmath>
using namespace std;
long double combi(int n,int m)
{
     if(n==m)
     return 1;
     else if(m==0)
     return 1;
     else return (n*combi(n-1,m-1))/m;
}
int main()
{
    int m,n;
    while(cin>>n>>m)
    {
                    if(n==0 && m==0)
                    break;
                    unsigned long long ans;
                    ans=(unsigned long long)combi(n,m);
                    cout<<n<<" things taken "<<m<<" at a time is "<<ans<<" exactly."<<endl;
    }
    system("pause");
}