hihocoder-Weekly221-Push Button II

hihocoder-Weekly221-Push Button II 

题目1 : Push Button II

时间限制:10000ms

单点时限:1000ms

内存限制:256MB

描述

There are N buttons on the console. Each button needs to be pushed exactly once. Each time you may push several buttons simultaneously.

Assume there are 4 buttons. You can first push button 1 and button 3 at one time, then push button 2 and button 4 at one time. It can be represented as a string "13-24". Other pushing way may be "1-2-4-3", "23-14" or "1234". Note that "23-41" is the same as "23-14".

Given the number N your task is to find the number of different valid pushing ways.

输入

An integer N. (1 <= N <= 1000)

输出

Output the number of different pushing ways. The answer would be very large so you only need to output the answer modulo 1000000007.

样例输入

3

样例输出

13

简单的DP问题 

将 dp[i][j] 定义为 i 个button 分为 j 个顺序的种类， 那么：

  dp[i][j] 可以由前一个状态转移得到， dp[i][j] = j * dp[i-1][j] + j * dp[i][j-1] 

其中 j * dp[i-1][j] 表示为 将第i个放入 含有1- (i-1) 的j 堆中，其中 i 可以放入j 中任意一组，种类为 j 

其中 j * dp[i][j-1] 表示为 第i个单独出来，形成第 j堆，那么整个新堆可以插入上面的次序中，一种j种插入方式。 

#include <cstdio>  
#include <cstring> 
#include <cstdlib>   

const int MAXN = 1000 + 10;  
const long long mod = 1000000007; 

int n; 
long long dp[MAXN][MAXN];  

int main(){ 

    while(scanf("%d", &n) != EOF)
    {
    	memset(dp, 0, sizeof(dp)); 
    	dp[1][1] = 1; 
    	for(int i=2; i<=n; ++i)
    	{ 
    		for(int j=1; j<=i; ++j)
    		{
    			dp[i][j] = (j * dp[i-1][j] + j*dp[i-1][j-1]) % mod; 
    		}
    	}  

    	long long ans = 0; 
    	for(int i=1; i<=n; ++i){
    		ans += dp[n][i];   
    	}
    	printf("%lld
", ans%mod );
    }
    return 0; 
}