计数训练之一

https://www.luogu.org/problem/P2606

不知道为什么这道题在数位dp

分析;又是一个与排列有关的计数题,

P(i)>P(i/2)这个条件很重要啊

也有P(2*i)>P(i),

也有P(2*i+1)>P(i)

像这种下标二倍的关系就要和二叉树考虑在一起

二叉树:i左儿子就是2*i,i的右儿子就是2*i+1

而且这颗二叉树是个完全二叉树

首先计算出i个节点的完全二叉树中,

根节点的左子树包含的节点数l,右子树包含的节点数r

首先,根节点的值必须为最小值。再考虑剩下的i-1个节点。

很容易想到,

可以在这i-1个节点中选出l个节点作为左子树,剩下的r个节点作为右子树。所以得出转移:

f[i]=C(i-1,l)×f[l]×f[r]

code by std(代码比较难看但真的很清晰):

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
inline int read() {
    int res = 0; bool bo = 0; char c;
    while (((c = getchar()) < '0' || c > '9') && c != '-');
    if (c == '-') bo = 1; else res = c - 48;
    while ((c = getchar()) >= '0' && c <= '9')
        res = (res << 3) + (res << 1) + (c - 48);
    return bo ? ~res + 1 : res;
}
const int N = 1e6 + 5;
int n, PYZ, f[N], fac[N], Log[N], inv[N];
int qpow(int a, int b) {
    int res = 1;
    while (b) {
        if (b & 1) res = 1ll * res * a % PYZ;
        a = 1ll * a * a % PYZ;
        b >>= 1;
    }
    return res;
}
int C(int x, int y) {
    if (!y) return 1;
    int u = C(x / PYZ, y / PYZ), v = x % PYZ, w = y % PYZ, z;
    if (v < w) z = 0;
    else z = 1ll * (1ll * fac[v] * inv[w] % PYZ) * inv[v - w] % PYZ;
    return 1ll * u * z % PYZ;
}
int main() {
    int i, kx, l = 1, r = 1; n = read(); PYZ = read();
    fac[0] = 1; Log[0] = -1;
    for (i = 1; i <= n; i++) fac[i] = 1ll * fac[i - 1] * i % PYZ,
        Log[i] = Log[i >> 1] + 1;
    kx = min(PYZ - 1, n); inv[kx] = qpow(fac[kx], PYZ - 2);
    for (i = kx - 1; i >= 0; i--) inv[i] = 1ll * inv[i + 1] * (i + 1) % PYZ;
    f[1] = f[2] = 1; f[3] = 2;
    for (i = 4; i <= n; i++) {
        if (i - (1 << Log[i]) + 1 <= (1 << Log[i] - 1)) l++;
        else r++;
        f[i] = 1ll * (1ll * C(i - 1, l) * f[l] % PYZ) * f[r] % PYZ;
    }
    printf("%d\n", f[n]);
    return 0;
}
posted @ 2019-10-13 08:36  wzx_believer  阅读(103)  评论(0编辑  收藏  举报