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 setT satisfies the following condition:

• T is a subset of S
• |T| = m
• T does not contain continuous numbers, that is to say x andx+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 1 115 2 11

Sample Output

56

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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<iostream>using namespace std;#define ll long longll 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<<ans<<endl;}return 0;}