zoj How Many Sets II 3557 （组合数学&&转换）好题

How Many Sets II

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

//题意：

给一个集合，一共n个元素，从中选取m个元素，满足选出的元素中没有相邻的元素，一共有多少种选法(结果对p取模1 <= p <= 10^9) 
思路：用插板法求出组合数。既然是从n个数中选择m个数，那么剩下的数为n-m，那么可以产生n-m+1个空，这道题就变成了把m个数插到这n-m+1个空中有多少种方法，即C(n-m+1,m)%p

#include
#include
#include
#define ll long long
using namespace std;
int n,m,p;
ll ksm(ll x,ll y)//快速幂 
{
	ll ans=1;
	while(y)
	{
		if(y&1)
			ans=(ans*x)%p;
		x*=x;
		x%=p;
		y>>=1;
	}
	return ans%p;
}
ll C(int n,int m)
{
	int i;
	ll sum1=1,sum2=1;
	for(i=1;i<=m;i++)
	{
		sum1=(sum1*(n-i+1))%p;
		sum2=(sum2*i)%p;
	}
	sum1=(sum1*ksm(sum2,p-2))%p;
	return sum1;
}
void solve(int n,int m)
{
	ll ans=1;
	while(n&&m&&ans)
	{
		ans=(ans*C(n%p,m%p))%p;
		n/=p;
		m/=p;
	}
	printf("%lld\n",ans);
}
int main()
{
	while(scanf("%d%d%d",&n,&m,&p)!=EOF)
	{
		n=n-m+1;
		if(nn-m)
			m=n-m;
		solve(n,m);
	}
	return 0;
}