You have a full binary tree having infinite levels.
Each node has an initial value. If a node has value x, then its left child has value 2·x and its right child has value 2·x + 1.
The value of the root is 1.
You need to answer Q queries.
There are 3 types of queries:
- Cyclically shift the values of all nodes on the same level as node with value X by K units. (The values/nodes of any other level are not affected).
- Cyclically shift the nodes on the same level as node with value X by K units. (The subtrees of these nodes will move along with them).
- Print the value of every node encountered on the simple path from the node with value X to the root.
Positive K implies right cyclic shift and negative K implies left cyclic shift.
It is guaranteed that atleast one type 3 query is present.
The first line contains a single integer Q (1 ≤ Q ≤ 105).
Then Q queries follow, one per line:
- Queries of type 1 and 2 have the following format: T X K (1 ≤ T ≤ 2; 1 ≤ X ≤ 1018; 0 ≤ |K| ≤ 1018), where T is type of the query.
- Queries of type 3 have the following format: 3 X (1 ≤ X ≤ 1018).
For each query of type 3, print the values of all nodes encountered in descending order.
5 3 12 1 2 1 3 12 2 4 -1 3 8
12 6 3 1 12 6 2 1 8 4 2 1
5 3 14 1 5 -3 3 14 1 3 1 3 14
14 7 3 1 14 6 3 1 14 6 2 1
Following are the images of the first 4 levels of the tree in the first test case:
Original:
After query 1 2 1:
After query 2 4 -1:
题意:三种操作
1.把值为x的那一层节点全部往右移k,子树不动
2.把值为x的那一层节点全部往右移k,子树一起走
3.问从节点编号为12的节点到根节点1的路径
思路:
核心:记录下每个深度往右移动多少次
模拟题意即可
注意位运算1LL
#include<bits/stdc++.h> #define PI acos(-1.0) using namespace std; typedef long long ll; const int N=1e5+5; const int MOD=1e9+7; const int INF=0x3f3f3f3f; int level(__int64 x){//获得值为x节点的深度 // cout <<"x="<<x<<endl; int cnt=0; while(x){ x/=2; cnt++; } return cnt; } __int64 sum[70]; __int64 shift[70]; int main(void){ ll q,x,k,op; for(int i=1;i<=64;i++) sum[i]=1ll<<(i-1); // cout<<sum[60]<<endl; cin >>q; for(ll i=1;i<=q;i++){ scanf("%I64d",&op); if(op==1){ scanf("%I64d%I64d",&x,&k); int dep=level(x); if(k>0) shift[dep]+=k; else shift[dep]+= sum[dep] +k;//左移k等价右移sum[dep]+k shift[dep]=(shift[dep]%sum[dep]+sum[dep])%sum[dep]; } else if(op==2){ scanf("%I64d%I64d",&x,&k); int dep=level(x); ll tot=1; for(;dep<64;dep++){//相对dep的深度决定要移动的个数 if(k>0) shift[dep]+=k%sum[dep]*tot%sum[dep]; else shift[dep]+= (sum[dep]+k)%sum[dep]*tot%sum[dep]; shift[dep]=(shift[dep]%sum[dep]+sum[dep])%sum[dep]; tot*=2; } } else{ scanf("%I64d",&x); int dep=level(x); for(int i=dep;i>=1;i--){ printf("%I64d ",x); x+=shift[i];//先求出当前x移动到的位置 if(x>= (1ll<<i)) x-=(1ll<<(i-1)); x/=2;//需要的x x+=(sum[i-1]-shift[i-1]);//哪一个编号移动到需要的x if(x>= (1ll<<(i-1))) x-=(1ll<<(i-2)); } puts(""); } } return 0; }