POJ 1942(组合数学)

题目链接：http://poj.org/status

题意：简单的来讲，就是求C(n,m）.

但是这题有毒..TLE2发也是醉了..

必须循环较小的一个，否则就会TLE。。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
using namespace std;

const int INF=0x3f3f3f3f;
const int Hashmaxn=100000;
const int maxn=10010;

typedef unsigned long long ULL;
ULL n,m;

ULL C(ULL n,ULL m){
	ULL res=1;
	for(ULL i=m+1,j=1;i<=n;i++,j++){
		res=res*i/j;
	}
	return res;
}

int main(){
#ifndef ONLINE_JUDGE
	freopen("test.in","r",stdin);
	freopen("test.out","w",stdout);
#endif
	while(cin>>n>>m){
		if(!n&&!m) break;
		if(n>m)
			cout<<C(n+m,n)<<endl;
		else
			cout<<C(n+m,m)<<endl;
	}
	return 0;
}