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).
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.
A single line contains a nonnegative integer, denoting the answer.
Sample Input
2
5 2
5 3
Sample Output
7
6
Source
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
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 #define MP make_pair 6 #define PB push_back 7 typedef long long LL; 8 typedef pair<int,int> PII; 9 const double eps=1e-8; 10 const double pi=acos(-1.0); 11 const int K=1e6+7; 12 const int mod=1e9+7; 13 14 LL pp[100],sum[100]; 15 LL go(LL x,LL y) 16 { 17 if(x==0) return 0; 18 LL cnt=1,ret=1; 19 if(y%2==0) return x; 20 while(1) 21 { 22 if(cnt==x) return ret; 23 cnt=cnt*y+1; 24 ret^=cnt; 25 } 26 } 27 LL dfs(LL n,LL k,int d) 28 { 29 if(n==1) return 1; 30 LL pos=n-1,ret; 31 for(int i=1;i<=d-2;i++) pos=(pos-1)/k; 32 ret=dfs(n-1-(pos-1)*sum[d-1]-(k-pos)*sum[d-2],k,d-1); 33 if((pos-1)&1) ret^=go(sum[d-1],k); 34 if((k-pos)&1) ret^=go(sum[d-2],k); 35 return ret^n; 36 } 37 int main(void) 38 { 39 LL n,k; 40 int t;cin>>t; 41 sum[1]=pp[1]=1; 42 while(t--) 43 { 44 cin>>n>>k; 45 if(k==1) 46 { 47 LL ans; 48 if(n%4==0) ans=n; 49 else if(n%4==1) ans=1; 50 else if(n%4==2) ans=n+1; 51 else if(n%4==3) ans=0; 52 printf("%lld ",ans); 53 continue; 54 } 55 int d=1; 56 for(int i=2;sum[i-1]<n;i++,d++) 57 pp[i]=pp[i-1]*k,sum[i]=sum[i-1]+pp[i]; 58 cout<<dfs(n,k,d)<<endl; 59 } 60 return 0; 61 }