Permutation Counting

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1487    Accepted Submission(s): 754

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

题意：对于任一种N的排列A，定义它的E值为序列中满足A[i]>i的数的个数。给定N和K(K<=N<=1000)，问N的排列中E值为K的个数。

dp[i][j]表示前i个数的排列中E值为j的个数，所以当插入第i+1个数时，如果放在第i+1或满足条件的j个位置时，j不变，与其余i-j个位置上的数调换时j会+1。所以

dp[i+1][j] = dp[i+1][j] + (j + 1) * dp[i][j];

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

/*
ID: LinKArftc
PROG: 3664.cpp
LANG: C++
*/

#include <map>
#include <set>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <cstdio>
#include <string>
#include <utility>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define eps 1e-8
#define randin srand((unsigned int)time(NULL))
#define input freopen("input.txt","r",stdin)
#define debug(s) cout << "s = " << s << endl;
#define outstars cout << "*************" << endl;
const double PI = acos(-1.0);
const double e = exp(1.0);
const int inf = 0x3f3f3f3f;
const int INF = 0x7fffffff;
typedef long long ll;

const int maxn = 1010;
const int MOD = 1000000007;

ll dp[maxn][maxn];
int n, k;

void init() {
    dp[1][0] = 1;
    dp[1][1] = 0;
    for (int i = 1; i <= 1000; i ++) {
        for (int j = 0; j <= 1000; j ++) {
            dp[i+1][j] = (dp[i+1][j] % MOD + (j + 1) * dp[i][j] % MOD) % MOD;
            dp[i+1][j+1] = (dp[i+1][j+1] % MOD + (i - j) * dp[i][j] % MOD) % MOD;
        }
    }
}

int main() {

    init();
    while (~scanf("%d %d", &n, &k)) {
        printf("%d\n", dp[n][k]);
    }

    return 0;
}