ACM HDU 3664 Permutation Counting

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

 

http://acm.hdu.edu.cn/showproblem.php?pid=3664

題意：對於任一種N的排列A，定義它的E值爲序列中知足A[i]>i的數的個數。給定N和K(K<=N<=1000)，問N的排列中E值爲K的個數。

解法：簡單DP。dp[i][j]表示i個數的排列中E值爲j的個數。假設如今已有一個E值爲j的i的排列，對於新加入的一個數i+1,將其加入排列的方法有三：1)把它放最後，加入後E值不變    2)把它和一個知足A[k]>k的數交換，交換後E值不變       3)把它和一個不知足A[k]>k的數交換，交換後E值+1      根據這三種方法獲得轉移方程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;  }