ZOJ 4536 ZOJ Monthly, October 2011 I

How Many Sets II

---

Time Limit: 2 Seconds Memory Limit: 65536 KB

---

Given a set S = {1, 2, ..., n}, number m and p, your job is to count how many set T satisfies the following condition:

• T is a subset of S
• |T| = m
• T does not contain continuous numbers, that is to say x and x+1 can not both in T

Input

There are multiple cases, each contains 3 integers n ( 1 <= n <= 10⁹ ), m ( 0 <= m <= 10⁴, m <= n ) and p ( p is prime, 1 <= p <= 10⁹ ) in one line seperated by a single space, proceed to the end of file.

Output

Output the total number mod p.

Sample Input

5 1 11
5 2 11

Sample Output

5
6

这个题也是找规律。

规律其实很简单，就是一个组合数：C(n-m+1,m)

但是注意一下当n-m+1<m的时候输出0。

当然，组合数取余，自然要用到lucas定理咯~

代码：

#include<stdio.h>
#include<string.h>
#include<iostream>

using namespace std;

typedef long long ll;

ll power(ll p,ll n,ll m)
{
	ll sq=1;
	while(n)
	{
		if(n%2==1)
			sq=(sq%m)*(p%m)%m;
		p=(p%m)*(p%m)%m;
		n=n/2;
	}
	return sq%m;
}

ll C(ll n,ll r,ll p)
{
	ll i,res=1,t;
	for(i=1;i<=r;i++)
	{
		res=(res%p)*((n-i+1)%p)%p;
		t=power(i,p-2,p);
		res=(res%p)*(t%p)%p;
	}
	return res;
}

ll lucas(ll n,ll r,ll p)
{
	if(n<r)
		return 0;
	return C(n,r,p);
}

int main()
{
	ll n,m,p,res,x,y;
	while(cin>>n>>m>>p)
	{
		if(n==0&&m==0&&p==0)
			break;
		res=1;
		x=n-m+1;
		y=m;
		if(x<y)
		{
			cout<<0<<endl;
			continue;
		}
		while(x&&y)
		{
			res=res*lucas(x%p,y%p,p)%p;
			if(res==0)
				break;
			x=x/p;
			y=y/p;
		}
		cout<<res<<endl;
	}
	return 0;
}