序列的计算 (图解

1.前言

被初一巨佬吊打

2.题解

1.状态转移方程

$dp[i][j]=dp[i-1][j]\ast (i-j)+dp[i-1][j]\ast (j+1)$

2.推导

分两种情况

①：

②:

综上所述：
$dp[i+1][j]=dp[i][j]\ast (j+1)+dp[i][j-1]\ast (i-j+1)$

$\Rightarrow dp[i][j]=dp[i-1][j]\ast (j+1)+dp[i-1][j-1]\ast (i-j)$

3.代码

代码${}_{极}$短，推导难想

#include <queue>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
using namespace std;

template <typename T> void read (T &x) {x = 0; T f = 1;char tem = getchar ();while (tem < '0' || tem > '9') {if (tem == '-') f = -1;tem = getchar ();}while (tem >= '0' && tem <= '9') {x = (x << 1) + (x << 3) + tem - '0';tem = getchar ();}x *= f;}
template <typename T> void write (T x) {if (x < 0) {x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');}
template <typename T> T Max (T x, T y) { return x > y ? x : y; }
template <typename T> T Min (T x, T y) { return x < y ? x : y; }
template <typename T> T Abs (T x) { return x > 0 ? x : -x; }

const int Maxn = 1005;
const LL Mod = 1e9 + 7;

int n, k;
LL dp[Maxn][Maxn];

int main () {
    for (int i = 1; i <= 1000; i++) {
        dp[i][0] = 1;
        for (int j = 1; j < i; j++) {
            dp[i][j] = (i - j) * dp[i - 1][j - 1] + (j + 1) * dp[i - 1][j];
            dp[i][j] %= Mod;
        }
    }//预处理 (打表)
    while (scanf ("%d %d", &n, &k) != EOF) {
        printf ("%lld\n", dp[n][k]);
    }
	return 0;
}