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


[align=left]Problem Description[/align]
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.
 

[align=left]Input[/align]
There are several test cases, and one line for each case, which contains two integers, N and k. (1 <= N <= 1000, 0 <= k <= N).

 

[align=left]Output[/align]
Output one line for each case. For the answer may be quite huge, you need to output the answer module 1,000,000,007.
 

[align=left]Sample Input[/align]

3 0
3 1

 

[align=left]Sample Output[/align]

1
4

HintThere 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}

 

[align=left]Source[/align]
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
[k + 1]);
}
}