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
d p [ i ] [ j ] = d p [ i − 1 ] [ j ] ∗ ( i − j ) + d p [ i − 1 ] [ j ] ∗ ( j + 1 ) dp[i][j] = dp[i - 1][j] * (i - j) + dp[i - 1][j] * (j + 1) dp[i][j]=dp[i−1][j]∗(i−j)+dp[i−1][j]∗(j+1)
2.推导
分两种情况
①：
②:
综上所述：
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dp[i + 1][j] = dp[i][j] * (j + 1) + dp[i][j - 1] * (i - j + 1)
dp[i+1][j]=dp[i][j]∗(j+1)+dp[i][j−1]∗(i−j+1)
⇒ d p [ i ] [ j ] = d p [ i − 1 ] [ j ] ∗ ( j + 1 ) + d p [ i − 1 ] [ j − 1 ] ∗ ( i − j ) \Rightarrow dp[i][j] = dp[i - 1][j] * (j + 1) + dp[i - 1][j - 1] * (i - j) ⇒dp[i][j]=dp[i−1][j]∗(j+1)+dp[i−1][j−1]∗(i−j)
3.代码
代码 极 _极 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;
}
