hdu6121 Build a tree

地址：http://acm.split.hdu.edu.cn/showproblem.php?pid=6121

题面:

Build a tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1240    Accepted Submission(s): 509

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

2
5 2
5 3

 

Sample Output

7
6

 

Source

2017 Multi-University Training Contest - Team 7

 

思路：

　　画了几个图后，会发现除了n-1号节点所在的链上需要特殊计算，其他字数都是满k叉树

　　所以递归下去计算即可，但对于1叉树要特判

 #include <bits/stdc++.h>

 using namespace std;

 #define MP make_pair
 #define PB push_back
 typedef long long LL;
 typedef pair<int,int> PII;
 const double eps=1e-;
 const double pi=acos(-1.0);
 const int K=1e6+;
 const int mod=1e9+;

 LL pp[],sum[];
 LL go(LL x,LL y)
 {
     if(x==) return ;
     LL cnt=,ret=;
     if(y%==) return x;
     while()
     {
         if(cnt==x) return ret;
         cnt=cnt*y+;
         ret^=cnt;
     }
 }
 LL dfs(LL n,LL k,int d)
 {
     if(n==) return ;
     LL pos=n-,ret;
     for(int i=;i<=d-;i++) pos=(pos-)/k;
     ret=dfs(n--(pos-)*sum[d-]-(k-pos)*sum[d-],k,d-);
     if((pos-)&) ret^=go(sum[d-],k);
     if((k-pos)&) ret^=go(sum[d-],k);
     return ret^n;
 }
 int main(void)
 {
     LL n,k;
     int t;cin>>t;
     sum[]=pp[]=;
     while(t--)
     {
         cin>>n>>k;
         if(k==)
         {
             LL ans;
             if(n%==) ans=n;
             else if(n%==) ans=;
             else if(n%==) ans=n+;
             else if(n%==) ans=;
             printf("%lld\n",ans);
             continue;
         }
         int d=;
         for(int i=;sum[i-]<n;i++,d++)
             pp[i]=pp[i-]*k,sum[i]=sum[i-]+pp[i];
         cout<<dfs(n,k,d)<<endl;
     }
     return ;
 }