LightOJ 1067 - Combinations （Lucas定理）

1067 - Combinations

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Time Limit: 2 second(s)

Memory Limit: 32 MB

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

For example, say there are 4 items; you want to take 2 ofthem. 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

 

题意：看样例就能看出题意了，C(n,m)%1000003

思路：1000003是素数，符合Lucas定理

ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 1010000
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
using namespace std;
LL gcd(LL a,LL b){return b?gcd(b,a%b):a;}
LL lcm(LL a,LL b){return a/gcd(a,b)*b;}
LL powmod(LL a,LL b,LL MOD){LL ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
//head
LL f[MAXN];
void findinit(LL p)
{
	f[0]=1;
	for(int i=1;i<=p;i++)
	f[i]=(f[i-1]*i)%p;
}
LL lucas(LL n,LL m,LL p)
{
	LL ans=1;
	while(n&&m)
	{
		LL a=n%p,b=m%p;
		if(a<b)
		return 0;
		ans=(ans*f[a]*powmod(f[b]*f[a-b]%p,p-2,p))%p;
		n/=p;m/=p;
	}
	return ans;
}
int main()
{
    findinit(1000003);
	LL n,m;
	int t,cas=0;
	scanf("%d",&t); 
	while(t--)
	{
		scanf("%lld%lld",&n,&m);
		LL ans=lucas(n,m,1000003);
		printf("Case %d: ",++cas);
		printf("%lld\n",ans);
	}
	return 0;
}