Hdu6121 Build a tree（2017多校第7场）

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本文探讨了一种特殊的完全k叉树构建方法，并通过递归算法求解该树中每个子树大小的异或和问题。特别针对k为奇数和偶数的情况进行了讨论。

Build a tree

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 556 Accepted Submission(s): 178

Problem Description

HazelFan wants to build a rooted tree. The tree has

n nodes labeled

0 to

n−1 , and the father of the node labeled

i is the node labeled

⌊i−1k⌋ . HazelFan wonders the size of every subtree, and you just need to tell him the XOR value of these answers.

Input

The first line contains a positive integer

T(1≤T≤5) , denoting the number of test cases.
For each test case:
A single line contains two positive integers

n,k(1≤n,k≤1018) .

Output

For each test case:
A single line contains a nonnegative integer, denoting the answer.

Sample Input

Sample Output

7
6

Source

2017 Multi-University Training Contest - Team 7

————————————————————————————————————

有一棵 $n$ 个点的有根树，标号为 $0$ 到 $n - 1$ ， $i$ 号点的父亲是 $\lfloor\frac{i-1}{k}\rfloor$ 号点，求所有子树大小的异或和。

这是一棵完全k叉树，考虑根的所有孩子，最多只有一个不是满k叉树，对这个孩子进行递归处理即可，剩下的可以直接算出来。k>1时只有log层，直接做就到底就好了，k=1时要特判

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <map>
#include <set>
#include <algorithm>
#include <complex>
#include <vector>
#include <bitset>
#include <stack>
#include <queue>
#include <unordered_map>
#include <functional>

using namespace std;

#define LL long long
const int INF=0x3f3f3f3f;

LL n,k;
LL ans;

LL llpow(LL x, LL y)
{
    LL ans = 1;
    while(y >= 1)
    {
        if(y & 1)
        {
            ans *= x;
        }
        x *= x;
        y >>= 1;
    }
    return ans;
}

void fff(LL x)
{
    if(x < 1)
        return;
    ans ^= x;
    if(k % 2 == 0)
    {
        LL nn = x, p = 1, cs = 1;
        nn--;
        while(nn - p * k >= 0) p *= k, nn -= p, cs++;
        if(nn == 0)
        {
            return;
        }
        LL t = llpow(k, cs - 1);
        LL tt = nn / t;
        if(nn % t == 0)
        {
            if(tt % 2 != 0)
            {
                fff((x - nn - 1) / k + t);
                fff((x - nn - 1) / k);
            }
        }
        else
        {
            if(tt % 2 != 0)
            {
                fff((x - nn - 1) / k + t);
                fff((x - nn - 1) / k + nn % t);
            }
            else
            {
                fff((x - nn - 1) / k + nn % t);
                fff((x - nn - 1) / k);
            }
        }
    }
    else
    {
        LL nn = x, p = 1, cs = 1;
        nn--;
        while(nn - p * k >= 0) p *= k, nn -= p, cs++;
        if(nn == 0)
        {
            fff((x - 1) / k);
            return;
        }
        LL t = llpow(k, cs - 1);
        LL tt = nn / t;
        if(nn % t == 0)
        {
            if(tt % 2 != 0)
            {
                fff((x - nn - 1) / k + t);
            }
            else
            {
                fff((x - nn - 1) / k);
            }
        }
        else
        {
            if(tt % 2 != 0)
            {
                fff((x - nn - 1) / k);
                fff((x - nn - 1) / k + t);
                fff((x - nn - 1) / k + nn % t);
            }
            else
            {
                fff((x - nn - 1) / k + nn % t);
            }
        }
    }
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%lld %lld",&n,&k);
        if(k==1)
        {
            if(n%4==0) printf("%lld\n",n);
            else if(n%4==1) printf("1\n");
            else if(n%4==2) printf("%lld\n",n+1LL);
            else printf("0\n");
            continue;
        }
        ans = 0;
        fff(n);
        printf("%lld\n", ans);
    }
    return 0;
}