每日一题之 hiho221 Push Button II (计数dp)

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

思路：

统计方案数的题目可以用dp来做，不妨我们设dp[i][j]表示 数字 1～i 分成 j 组一共有多少种方案。那么考虑 dp[i+1][j+1]等于多少？ 在之前由于递推求得了 dp[i][j] 了，那么

1. 一种情况是1～i 的数分成 j 组，然后第 i+1个数 独立成为一组 ，相当于在分好的 j 组中 插空就行了，j 个组 一共 j+1 个空，所以这种情况的方案数是$dp[i][j]\ast (j+1)$
2. 另外一种情况是 1～i的数已经被分成了 j+1 个组，那么第 i+1 这个数，随便放到这 j+1中的任意一个组就行了，那么这种情况的方案数是 $d[i][j+1]\ast (j+1)$

综上就可以得出递推式：$dp[i][j]=dp[i-1][j-1]\ast j+dp[i-1][j]\ast j$

#include <cstdio>
#include <iostream>
#include <cstring>

using namespace std;

const int maxn = 1e3+5;
const int mod = 1e9+7;

typedef long long ll;

ll dp[maxn][maxn];

void solve(int n)
{
	memset(dp,0,sizeof(dp));
	for (int i = 1; i <= n; ++i)
		dp[i][1] = 1;

	for (int i = 2; i <= n; ++i) {
		for (int j = 1; j <= n; ++j) {
			dp[i][j] = (dp[i-1][j-1]*j)%mod + (dp[i-1][j]*j)%mod;
			dp[i][j] %= mod;
		}
	}
	ll res = 0;
	for (int i = 1; i <= n; ++i) {
		res = (res + dp[n][i])%mod;
	}
	cout << res%mod << endl;
}

int main()
{

	int n;
	cin >> n;
	solve(n);

	return 0;
}