Lightoj 题目1067 Combinations（lucas）

最新推荐文章于 2019-11-06 13:09:55 发布

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最新推荐文章于 2019-11-06 13:09:55 发布

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本文介绍了一种高效计算组合数并进行大数取余的方法。通过预处理阶乘及使用Lucas定理，能够在短时间内解决给定大量测试用例的问题。文章提供了完整的C++实现代码，展示了如何通过快速幂运算求逆元来简化复杂度。

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C - Combinations

Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu

Submit Status Practice LightOJ 1067

Description

Given n different objects, you want to take k of them. How many ways to can do it?

For example, say there are 4 items; you want to take 2 of them. So, you can do it 6 ways.

Take 1, 2

Take 1, 3

Take 1, 4

Take 2, 3

Take 2, 4

Take 3, 4

Input

Input starts with an integer T (≤ 2000), denoting the number of test cases.

Each test case contains two integers n (1 ≤ n ≤ 10⁶), k (0 ≤ k ≤ n).

Output

For each case, output the case number and the desired value. Since the result can be very large, you have to print the result modulo 1000003.

Sample Input

4 2

5 0

6 4

Sample Output

Case 1: 6

Case 2: 1

Case 3: 15

组合数取余

ac代码

611086	2015-11-09 14:27:54	1067 - Combinations	C++	0.028	9500	Accepted

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<stdlib.h>
#include<map>
#include<math.h>
#define LL long long
using namespace std;
LL fac[1000005];
void get_fact(LL p)
{
	fac[0]=1;
	for(int i=1;i<=p;i++)
		fac[i]=(fac[i-1]*i)%p;
}
LL qpow(LL a,LL b,LL mod)
{
	LL ans=1;
	while(b)
	{
		if(b&1)
			ans=(ans*a)%mod;
		a=(a*a)%mod;
		b>>=1;
	}
	return ans;
}
LL lucas(LL n,LL m,LL mod)
{
	LL ans=1;
	while(n&&m)
	{
		LL a=n%mod;
		LL b=m%mod;
		if(a<b)
			return 0;
		ans=(ans*fac[a]*qpow(fac[b]*fac[a-b]%mod,mod-2,mod))%mod;
		n/=mod;
		m/=mod;
	}
	return ans;
}
int main()
{
	int t,c=0;
	scanf("%d",&t);
	get_fact(1000003);
	while(t--)
	{
		LL n,m,p;
		scanf("%lld%lld",&n,&m);
		printf("Case %d: %lld\n",++c,lucas(n,m,1000003));
	}
}