UVA - 1485 Permutation Counting

Description

Given a permutation a₁, a₂,...a_N of{1, 2,..., N}, we define its E-value as the amount of elements where a_i >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 exactlyk.

Input

There are several test cases, and one line for each case, which contains two integers,N and k. (1N1000, 0kN).

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.

Explanation for the sample:

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}

Sample Input

30 31

Sample Output

1 4题意：给定一个1-n的排列，满足ai > i的下标i的个数称为此排列的E值，给定整数n和k，求E值恰好为k的排列个数。思路：先写出转移方程: dp[i][j] = (j+1)*dp[i-1][j] + (i-j)*dp[i-1][j-1]，我们拿第i个考虑，如果之前有dp[i-1][j]的话，拿它去和那些满足条件的下标换的话，因为第i个数是最大的，所以此时满足的个数是不变的；如果是dp[i-1][j-1]的话，那么我们去和那些((i-1)-(j-1))不满足的换的话满足的个数会+1。
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>typedef long long ll;using namespace std;const int maxn = 1010;const ll mod =  1000000007;ll dp[maxn][maxn];int n, k;void init() {for (int i = 1; i < maxn; i++) {dp[i][0] = 1;for (int j = 1; j < i; j++)dp[i][j] = ((j+1)*dp[i-1][j] + (i-j)*dp[i-1][j-1]) % mod;}}int main() {init();while (scanf("%d%d", &n, &k) != EOF) {printf("%lld\n", dp[n][k]);}return 0;}