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