【LightOJ】1067 - Combinations（Lucas & 逆元）

FishWang

发布于 2025-08-27 09:56:57

1980

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1067 - Combinations

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Time Limit: 2 second(s)	Memory Limit: 32 MB

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	Output for Sample Input
3 4 2 5 0 6 4	Case 1: 6 Case 2: 1 Case 3: 15

直接套用Lucas定理的公式就行了。具体见下一篇博客，写了个总结。

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define CLR(a,b) memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define LL long long
LL mod = 1000003;
LL fac[1000000+11] = {1,1};
void getfac()		//打一个阶乘表 
{
	for (int i = 2 ; i <= 1000000 ; i++)
		fac[i] = fac[i-1] * i % mod;
}
LL quick(LL n , LL m)		//求快速幂 
{
	LL ans = 1;
	n %= mod;
	while (m)
	{
		if (m & 1)
			ans = ans * n % mod;
		n = n * n % mod;
		m >>= 1;
	}
	return ans;
}
LL C(LL n , LL k)		//费马小定理求逆元 
{
	if (k > n)
		return 0;
	else
		return fac[n] * (quick(fac[k] * fac[n-k] % mod , mod - 2)) % mod;
}
LL Lucas(LL n,LL k)		//Lucas定理递归 
{
	if (k == 0)		//递归终止条件 
		return 1;
	else
		return C(n % mod , k % mod) * Lucas(n / mod , k / mod) % mod;
}
int main()
{
	getfac();
	LL n,k;
	int Case = 1;
	int u;
	scanf ("%d",&u);
	while (u--)
	{
		scanf ("%lld %lld",&n,&k);
		printf ("Case %d: %lld\n",Case++,Lucas(n,k));
	}
	return 0;
}

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原始发表：2025-08-26，如有侵权请联系 cloudcommunity@tencent.com 删除

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