HDU-3664-Permutation Counting

Permutation Counting

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1739 Accepted Submission(s): 918

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

题意：E条件：就是在下标为i处的数值ai大于i （i从1开始）给出数串的长度N 与k 问在当下数串长度下1，2，3。。。N 的数串排列中有多少个排列符合E条件

dp 。。。

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

即在i长度下每个排列的E条件数为j的排列数 = 当在i-1长度下有j个E条件的排列数 + 在i-1长度下有j个E条件数的排列数 * （新数来了后与符合条件的数交换仍然符合E条件）+ 在i-1长度下有j-1个符合E条件的数的排列数 * （i-j）【即把每个有j-1个符合的排列中不符合E条件的数与新来的第i个数交换符合条件的数会++】

也就是说把直接放后面的还有与符合条件数交换的还有不符合条件的数交换的所有情况都考虑在内

code。。。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MOD = 1000000007;
ll d[1010][1010]; 
int main()
{d[1][0]=1;for(int i=1;i<=1000;i++)d[1][i]=0;for(int i=2;i<=1000;i++){for(int j=0;j<=i;j++){d[i][j]=d[i-1][j]+d[i-1][j]*j+d[i-1][j-1]*(i-j);	d[i][j]%=MOD;}	} int n,k;while(cin>>n>>k){cout<<d[n][k]<<endl;}return 0;
}

注意结果取模并且dp数组要开ll

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