2022 Jiangsu Collegiate Programming Contest - J. Balanced Tree_a binary tree tt is called super balanced if tt is-优快云博客

本文链接：https://blog.youkuaiyun.com/Szyangc/article/details/127199797

本文介绍了一种高效算法，用于计算具有特定平衡条件的二叉树数量。通过递推公式结合对数运算，实现了对大规模数据的有效处理，特别适用于求解节点数量巨大时的平衡二叉树组合数目。

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L. Collecting Diamonds

题目描述

A binary tree T is called super balanced if T is empty or satisfies the following three conditions at the same time:

The left subtree of T is super balanced.
The right subtree of T is super balanced.
The number of nodes of the left subtree and the right subtree differs at most 1.
Please calculate the number of super balanced trees consisting of n nodes. Since the answer can be very large, just output the answer modulo 2^64.

输入描述

The first line contains a single integer T (1≤T≤106), indicating the number of test cases.

Each test case contains a single integer n (0≤n<264), indicating the number of nodes in the tree.

输出描述

For each test case, output an integer in a single line indicating the number of super balanced trees consisting of n nodes.

输入样例1：

2 2 3

输出样例1：

2 1

题意

求构造一棵使每个节点的左右子树的 size 之差小于等于 1 的方案数
对 2^64 取模

思路

设 f[x] 为有 x 个节点的方案树
特殊的 f[0]=1
i）若 x为偶数： $f [x] = f [x /2] * f [x /2 - 1] * 2$
ii）若 x为奇数： $f[x]=f[x/2]^2$
观察柿子可知都是以2的幂次在进行计算
故我们设 $g[x]=log_2(f[x])$
i）若 x为偶数： $g [x] = g [x /2] + g [x /2 - 1] + 1$
ii）若 x为奇数： $g [x] = 2 * g [x /2]$

但这题 x 有点大
时间空间不允许记忆化搜索
我们设 $g [n] = a * g [x] + b * g [x - 1] + c$
我们只需要根据当前 x 的奇偶更新 x , a ,b ,c 即可直至 x==1 可发现是log级的

最终状态为 $g [n] = a * g [1] + b * g [0] + c = c$

故答案为 2^c % 2^64
可发现当 c>=64 时 2 ^ c | 2 ^ 64 故取模后为 0 可以省几次快速幂
不省的话会有点T 需要加个快读快写优化一下

Code

#include<bits/stdc++.h>
using namespace std;
#define __T int csT;scanf("%d",&csT);while(csT--)
const int mod=1e9+7;//2^64 ull自然溢出即可
const int maxn=1e5+3;

unsigned long long n,a,b,c;
unsigned long long qpow(unsigned long long dd,unsigned long long x)
{
    unsigned long long d=dd,sum=1;
    while(x>0)
    {
        if(x%2==1)sum=(sum*d);
        d=(d*d);
        x>>=1;
    }
    return sum;
}
inline void sol()
{
    scanf("%llu",&n);
    //g[n]=1*g[n]+0*g[n-1]+o*c;
    a=1;
    b=0;
    c=0;
    while(n>1)
    {
        if(n%2==0)
        {
            b=a+2*b;
            c=a+c;
            n/=2;
        }
        else
        {
            a=2*a+b;
            c=b+c;
            n=(n-1)/2;
        }
    }
    if(c>=64)puts("0");
    else printf("%llu\n", qpow(2,c));
}

int main()
{
    __T
        sol();
    return 0;
}