HDU 3664 Permutation Counting （DP）

Permutation Counting

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1606    Accepted Submission(s): 827

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

 

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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。

思路：dp[i][[j]表示前i个数中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);

注意取模，注意用__int64 !!!!!!!!

#include<stdio.h>#include<string.h>#include<algorithm>#define mod 1000000007using namespace std;__int64 dp[1010][1010];int main(){int n,k,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;}