| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 18457 | Accepted: 5633 |
Description
Input
Output
Sample Input
4 2 10 5 49 6 0 0
Sample Output
6 252 13983816
题意：求C(n,m)；
思路：这个是其中一种办法，就是连乘r个整商：C(n,k)=C(n,k-1)*(n-k+1)/k。时间复杂度O(n);
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#include <set>
#include <queue>
#include <stack>
#include <map>
using namespace std;
typedef long long LL;
LL work(LL n,LL m)
{
if(m>n/2) m=n-m;
LL a=1,b=1;
for(int i=1;i<=m;i++){
a*=n-i+1;
b*=i;
if(a%b==0){
a/=b;
b=1;
}
}
return a/b;
}
int main()
{
LL n,m;
while(~scanf("%lld %lld",&n,&m)){
if(!n&&!m) break;
printf("%lld\n",work(n,m));
}
return 0;
}
POJ 2249-Binomial Showdown(排列组合计数)
原文：http://blog.csdn.net/u013486414/article/details/44260547