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

題意  ：E條件：就是在下标為i處的數值ai大于i （i從1開始） 給出數串的長度N 與k 問在當下數串長度下1，2，3。。。N 的數串排列中 有多少個排列符合E條件

dp 。。。    d[i][j]=d[i-1][j]+d[i-1][j]*j+d[i-1][j-1]*(i-j); 即 在i長度下每個排列的E條件數為j的排列數  =  當在i-1長度下  有j個E條件的排列數 + 在i-1長度下有j個E條件數的排列數 * （新數來了後與符合條件的數交換 仍然符合E條件）+ 在i-1長度下  有j-1個符合E條件的數的排列數 * （i-j）【即把 每個有j-1個符合的排列中不符合E條件的數與新來的第i個數交換 符合條件的數會++】 也就是說 把直接放後面的 還有與符合條件數交換的 還有不符合條件的數交換的所有情況都考慮在内

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;
}
           

注意結果取模 并且dp數組要開ll