POJ2249 Binomial Showdown

最新推荐文章于 2018-09-02 08:49:54 发布

-斯德哥尔摩-

最新推荐文章于 2018-09-02 08:49:54 发布

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分类专栏：数论

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本文介绍了一种高效计算组合数C(n,k)的方法，避免了直接使用公式带来的整数溢出问题。通过预先存储分子和分母，逐步消去最大公约数进行简化，最终利用long long类型变量计算出结果。

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Description

In how many ways can you choose k elements out of n elements, not taking order into account?
Write a program to compute this number.

Input

The input will contain one or more test cases.
Each test case consists of one line containing two integers n (n>=1) and k (0<=k<=n).
Input is terminated by two zeroes for n and k.

Output

For each test case, print one line containing the required number. This number will always fit into an integer, i.e. it will be less than 2³¹.
Warning: Don't underestimate the problem. The result will fit into an integer - but if all intermediate results arising during the computation will also fit into an integer depends on your algorithm. The test cases will go to the limit.

Sample Input

Sample Output

6
252
13983816

题目大意：求 C ( n , k )

普通的公式肯定会 WA 。

可以将每个分子分母存起来，约去 gcd ，再用 long long 计算答案。

附代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#define MAXN 10010
using namespace std;
long long n,k,a[MAXN],b[MAXN];
inline int read(){
	int date=0,w=1;char c=0;
	while(c<'0'||c>'9'){if(c=='-')w=-1;c=getchar();}
	while(c>='0'&&c<='9'){date=date*10+c-'0';c=getchar();}
	return date*w;
}
long long gcd(long long x,long long y){
	if(!y)return x;
	return gcd(y,x%y);
}
void work(){
	if(n-k<k)k=n-k;
	for(long long i=1;i<=k;i++){a[i]=i;b[i]=n-k+i;}
	for(int i=1;i<=k;i++)
	for(int j=2;j<=k;j++){
		if(a[j]==1)continue;
		long long g=gcd(b[i],a[j]);
		if(g!=1){b[i]/=g;a[j]/=g;}
		if(b[i]==1)break;
	}
	long long ans=1;
	for(int i=1;i<=k;i++)if(b[i]!=1)ans*=b[i];
	printf("%lld\n",ans);
}
int main(){
	while(1){
		n=read();k=read();
		if(n==0&&k==0)break;
		memset(a,0,sizeof(a));memset(b,0,sizeof(b));
		work();
	}
	return 0;
}