HDU - 3664 Permutation Counting 排列规律dp

Permutation Counting

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.

InputThere are several test cases, and one line for each case, which contains two integers, N and k. (1 <= N <= 1000, 0 <= k <= N).
OutputOutput 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}

数列1-n，可以随意排列组合，求恰有k个a[i]>i的排列个数。

一道找规律的dp。看数据范围就知道不能暴力求解，但可以用暴力找出n较小的几种小数列排列数，发现规律。类似杨辉三角，就像两数和靠拢，于是可以发现状态转移方程f[i][j]=(((i+1)*f[i][j-1])%MOD+((j-i)*f[i-1][j-1])%MOD)%MOD

#include<stdio.h>

#include<stdlib.h>

#include<string.h>

#include<string>

#include<math.h>

#include<queue>

#include<set>

#include<stack>

#include<algorithm>

#include<vector>

#include<iterator>

#define MAX 1005

#define INF 0x3f3f3f3f

#define MOD 1000000007

using namespace std;

typedef long long ll;

ll f[MAX][MAX];

int main()

{

    int n,m,i,j;

    for(i=;i<=;i++){

        f[][i]=;

    }

    for(i=;i<=;i++){

        for(j=i+;j<=;j++){

            f[i][j]=(((i+)*f[i][j-])%MOD+((j-i)*f[i-][j-])%MOD)%MOD;

        }

    }

    while(~scanf("%d%d",&n,&m)){

        if(n==m){

            printf("0\n");

            continue;

        }

        printf("%lld\n",f[m][n]);

    }

    return ;

}

巴特西

HDU - 3664 Permutation Counting 排列规律dp

Permutation Counting

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