HDU 3664 Permutation Counting （DP）

Permutation Counting

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 493    Accepted Submission(s): 258

Problem Description

Given a permutation a1, a2, … aN of {1, 2, …, N}, we define its E-value as the amount of elements where ai > i. For example, the E-value of permutation {1, 3, 2, 4} is 1, while the E-value of {4, 3, 2, 1} is 2. You are requested to find how many permutations of {1, 2, …, N} whose E-value is exactly k.

 

Input

There are several test cases, and one line for each case, which contains two integers, N and k. (1 <= N <= 1000, 0 <= k <= N).

 

Output

Output one line for each case. For the answer may be quite huge, you need to output the answer module 1,000,000,007.

 

Sample Input

3 0 3 1

 

Sample Output

1 4

Hint

There is only one permutation with E-value 0: {1,2,3}, and there are four permutations with E-value 1: {1,3,2}, {2,1,3}, {3,1,2}, {3,2,1}

 

Source

2010 Asia Regional Harbin

 

Recommend

lcy

题意：对于任一种N的排列A，定义它的E值为序列中满足A[i]>i的数的个数。给定N和K(K<=N<=1000)，问N的排列中E值为K的个数。

解法：简单DP。dp[i][j]表示i个数的排列中E值为j的个数。假设现在已有一个E值为j的i的排列，对于新加入的一个数i+1,将其加入排列的方法有三：1)把它放最后，加入后E值不变    2)把它和一个满足A[k]>k的数交换，交换后E值不变       3)把它和一个不满足A[k]>k的数交换，交换后E值+1      根据这三种方法得到转移方程dp[i][j] = dp[i - 1][j] + dp[i - 1][j] * j + dp[i - 1][j - 1] * (i - j);

拿到dp还是没有套路，没有方向的乱做。。看了题解之后觉得，这一看就是二维套路啊。。。也没对这进行好好分析。。唉。。要狠练dp了！

#include <iostream>#include <algorithm>#include <cstring>#include <cstdio>using namespace std;const int maxn = 1e3 + 5;const int Mod = 1e9 + 7;typedef long long ll;ll dp[maxn][maxn];int main(){    for(int i = 0; i < maxn; i++)        dp[i][0] = 1;    for(int i = 2; i < maxn; i++)        for(int j = 1; j <= i; j++)        {            dp[i][j] = (dp[i-1][j]+dp[i-1][j]*j+dp[i-1][j-1]*(i-j))%Mod;        }    int n, k;    while(~scanf("%d%d", &n, &k))    {        cout << dp[n][k] << endl;    }    return 0;}