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
 

 

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);

#include<stdio.h>
#include<iostream>
using namespace std;
const int MAXN=1001;
long long dp[MAXN][MAXN];
const long long MOD=1000000007;
int main()
{
int n,k;
int i,j;
for(i=1;i<=1000;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)!=EOF)
      printf("%I64d\n",dp[n][k]);
return 0;  
}