hdu 6121 Build a tree （图论）

Build a tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1204    Accepted Submission(s): 493

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

 

题意：有一棵n个点的有根数，根编号为0，i号节点的父亲是(i-1)/k号节点

求出所有子树大小的异或和

性质：对于一颗满k叉树而言，如果k是偶数，那么它的异或和就是树的大小

如果k是奇数，那么它的异或和就是任意一棵子树大小的异或和再异或树的大小

而这题的树也有几个性质：

①假设n足够大，那么根有k个儿子（废话）

②这k棵子树最多只有一棵不是满二叉树

那么假设k不等于1，二叉树的深度不会超过63

所以只要找到是哪棵子树不是满二叉树，以那棵子树作为整棵树求它的异或和，很显然子树也具有一样的性质，这样就可以递归求解，求出来之后直接异或其它所有满二叉树的大小就是答案，注意别忘记异或上n

还有k=1要特判

#include<cstring>#include<cstdio>#include<algorithm>#include<queue>#include<cmath>#include<set>using namespace std;typedef long long int ll;const int N = 2e5+5;ll ans=0,n,k;void solve(){ans=n;while(1)  //通过不断删树，使得最后的的非满子树变成一个2层的树{                 //每一次计算除掉那棵非满子树外的其他树的异或和if(n-1<=k)   //若树删减到只有两层的情况时{if((n-1)%2)ans^=1;  //ans已经异或过根节点了break;}ll now,x,res;now=x=res=1;     //now当前节点数总和，x表示当前层总的节点数while(0<=now+x*k&&now+x*k<n){x=x*k;now+=x;res^=now;}//now表示现在除最底层外其他节点的节点数，x表示now表示的层的节点数ll l,r;// l表示在那棵非满子树左边有多少棵子树，r同理l=(n-now-1)/x;r=k-1-l;if(k%2){if(l%2) ans^=res;if(r%2) ans^=res^(now-x);}else{if(l%2) ans^=now;if(r%2) ans^=(now-x);}n-=l*now+1+r*(now-x);  //将已经计算过的树删掉，只留下那个非满二叉树ans^=n;   //异或删完之后的根节点}}int main() {    int t;ll i;scanf("%d",&t);while(t--){scanf("%lld%lld",&n,&k);if(k==1){if(n%4==1) printf("%lld\n",1);else if(n%4==2) printf("%lld\n",n+1);else if(n%4==3) printf("%lld\n",0);else printf("%lld\n",n);}else{solve();printf("%lld\n",ans);}}return 0;}