HDU 3664 递推

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原文链接：http://www.cnblogs.com/zhangmingzhao/p/7229722.html

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Permutation Counting

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1794 Accepted Submission(s): 955

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}

和上一篇csu多校的差不多，替换i个位置(最后一个位置直接放)：

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。

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);

注意i和j从1开始，maxn是不能等到，越界

#include <bits/stdc++.h>
using namespace std;
const int maxn=1000+10;
long long dp[maxn][maxn];
const int mod = 1000000007;

int main()
{
    int n,k;
    int i,j;
    for(i=1;i<maxn;i++)
    {
        dp[i][0]=1;
        for(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;
    }
    while(~scanf("%d%d",&n,&k))
      printf("%I64d\n",dp[n][k]);
    return 0;
}