HDOJ 3664 Permutation Counting

Permutation Counting

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1407    Accepted Submission(s): 711

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} 
    
    
    
    
     
   
   
   
   

 

 

虽说是DP水题，但对于渣渣要说实在是不会。所以有幸拜访了Kuangbin大神的博客。。。。。

 

 

题意：给出数字n和k，问有多少个数列满足有k个数且这这些数比数字本身对应的下标的值大。

 

解题思路：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);

 

具体代码：

 

#include<stdio.h>
#define mod 1000000007
__int64 dp[1010][1010];//不能用int数组，数列的个数会很大 
int main()
{
	int n,k,i,j;
	for(i=1;i<=1000;i++)
	{
		dp[i][0]=1;
		for(j=1;j<i;j++)//不存在n与k值相等的数列 
		   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("%d\n",dp[n][k]);
	return 0;
}