HDU 3664 Permutation Counting 解题报告（递推）

Permutation Counting

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1446    Accepted Submission(s): 736

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 03 1

 

Sample Output

14
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

 

    解题报告：下午队伍一起做了一场哈工大的比赛，说起来很搞笑，三个人都没思路（真的……）。简单暴力，列出简单数据的结果，然后找规律……

    得到一个类似于杨辉三角的图形，然后三个人一起YY公式，然后就A了……

    正式正确的姿势不是这样的啊(╯‵□′)╯︵┴─┴

    赛后看了解题报告。dp[i][j]表示使用1~i这些数，使得k=j的情况数。那么递推公式为：

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

    简单来说就是新增一个数i时，如果把这个放在第i位，或者与前面出现的ax > x中的某个数交换，k保持不变；如果与ax <= x交换，则k+1。

    边界条件为dp[0][0] = 1，递推即可，没啥好说的……

#include <cmath>#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <string>#include <set>#include <map>#include <queue>#include <vector>#include <functional>using namespace std;typedef long long ll;typedef unsigned long long ull;#define ff(i, n) for(int i=0,END=(n);i<END;i++)#define fff(i, n, m) for(int i=(n),END=(m);i<=END;i++)#define dff(i, n, m) for(int i=(n),END=(m);i>=END;i--)#define mid ((l+r)/2)#define bit(n) (1ll<<(n))#define clr(a, b) memset(a, b, sizeof(a))void work();int main() {    work();    return 0;}/**************************Beautiful GEGE**********************************/const int mod = 1e9 + 7;const int maxn = 1111;ll dp[maxn][maxn];void work() {    dp[0][1] = 1;    fff (i, 1, 1000) fff (j, 1, i) {        dp[i][j] = (dp[i - 1][j] * j + dp[i - 1][j - 1] * (i - j + 1)) % mod;    }    int n, k;    while(scanf("%d%d", &n, &k) == 2) {        printf("%lld\n", dp[n][k + 1]);    }}