HDU 3664 Permutation Counting

Permutation Counting

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1330    Accepted Submission(s): 669

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 03 1

 

Sample Output

14
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

解题思路：给你1到n的n个数的全排列，求使得a[i]>i的数恰有k个方案的个数

令dp[i][j]表示i个数有j个a[i]>i的方案数，对于dp[i-1][j-1]，最后一个数我们可以使它与a[i]<=i的任意一个数交换，共有(i-1-(j-1)) 种可能，对于dp[i-1][j],它要么直接放在最后一位，要么与a[i]>i的任意一个数交换，共有(j+1)种可能，所以

dp[i][j]=dp[i-1][j-1]*(i-j)+dp[i-1][j]*(j+1);

#include<cstdio>#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;  }