Hdu6121 Build a tree（2017多校第7场）

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

25 25 3

 

Sample Output

76

 

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⌊​k​​i−1​​⌋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 longconst 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;}