ZOJ 3557 How Many Sets II

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

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Author: QU, Zhe
Contest: ZOJ Monthly, October 2011

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，这样的选法一共有多少种？

说的高大上就是隔板法。其实就是。和正常做法没什么差别。

要从N个元素里面取M个元素。就是C(N,M)

应为题目要求没有相邻的元素，所以假设你已经取出了M个球，在取球的过程中，有M-1个球是肯定不能取得。

所以能取得球就是n-(m-1);

所以就是C(N-(M-1),M).

应为这里的数据范围特别大，所以要用到Lucas

#include
using namespace std;
#define ll long long
ll n,m,p;

long long make_pow(long long x,long long y,long long mod)
{
	long long res =1;
	while(y>0)
	{
		if(y&1)res = res*x%mod;
		x=x*x%mod;
		y>>=1; 
	} 
	return res;
}

long long C(long long x,long long y)  
 {  
     if(y>x)  
         return 0;  
     else       
	 {  
         long long a,b,ans=1;           
		 for(long long i=1;i<=y;i++)  
         {  
             a=(x+i-y)%p;  
             b=i%p;  
             ans=ans*(a*make_pow(b,p-2,p)%p)%p;  
         }  
         return ans;  
     }  
 }  


long long Lucas(long long x, long long y)  {  
    if(y==0)  return 1;  
    else      return (C(x%p,y%p)*Lucas(x/p,y/p))%p;  
}  
int main()
{
	
	while(cin >> n >> m >> p)
	{
		ll ans;
		ans = Lucas(n-m+1,m)%p;
		cout<