CodeForces - 960D Full Binary Tree Queries

﹏ヽ暗。殇╰゛Y 2022-05-08 00:40 129阅读 0赞

## 题目：[点击打开链接][Link 1] ##

## 题意：无限长度的二叉树,每次操作1把包含数x的那一层整体移动k个位置，正往右，负往左，操作2把包含数x的那一层带着子树移动k个位置.操作3打印从x到根沿线的所有数。 ##

## 分析：对于每一层我们可以记录旋转了多少次,因为最多有60+层.对于操作2它的子树就分别移动2\*k次,4\*k次,8\*k...，对于查询操作,我们可以先找到x的位置，然后依次除以2就是上层的对应位置,根据旋转次数打印相应数字即可.注意一下细节就好了. ##

## 代码： ##

#pragma comment(linker, "/STACK:102400000,102400000")
    #include<unordered_map>
    #include<unordered_set>
    #include<algorithm>
    #include<iostream>
    #include<fstream>
    #include<complex>
    #include<cstdlib>
    #include<cstring>
    #include<cassert>
    #include<iomanip>
    #include<string>
    #include<cstdio>
    #include<bitset>
    #include<vector>
    #include<cctype>
    #include<cmath>
    #include<ctime>
    #include<stack>
    #include<queue>
    #include<deque>
    #include<list>
    #include<set>
    #include<map>
    using namespace std;
    #define pt(a) cout<<a<<endl
    #define debug test
    #define mst(ss,b) memset((ss),(b),sizeof(ss))
    #define rep(i,a,n) for (int i=a;i<=n;i++)
    #define per(i,a,n) for (int i=n-1;i>=a;i--)
    #define fi first
    #define se second
    #define ll long long
    #define ull unsigned long long
    #define pb push_back
    #define mp make_pair
    #define inf 0x3f3f3f3f
    #define eps 1e-10
    #define PI acos(-1.0)
    const ll mod = 1e9+7;
    const int N = 1e6+10;
    
    ll qp(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
    int to[4][2]={
       {-1,0},{1,0},{0,-1},{0,1}};
    
    ll q,t,x,k,dep[100],pw[62];
    
    int main() {
        ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
        for(ll i=1,j=2;i<62;i++,j<<=1) pw[i]=j;
        pw[0]=1;
        cin>>q;
        while(q--) {
            cin>>t>>x;
            if(t==3) {
                ll u=log2(x);
                ll now=pw[u]+(x-pw[u]+dep[u])%pw[u];///在树上的位置
                cout<<x<<" ";
                now/=2;
                for(ll i=u-1;i>=0;i--,now/=2) {
                    if(now-dep[i]>=pw[i]) cout<<now-dep[i]<<" ";///分情况讨论
                    else cout<<(now-dep[i]+pw[i])%pw[i+1]<<" ";///当前位置上的数字
                }
                cout<<endl;
            }else {
                cin>>k;
                ll u=log2(x),mdp,l;
                if(t==1) mdp=u;
                else mdp=60;
                for(ll i=u,l=(k%pw[u]+pw[u])%pw[u];i<=mdp;i++,l<<=1)///按规律更新
                    dep[i]=(dep[i]+l)%pw[i];
            }
        }
        return 0;
    }

[Link 1]: http://codeforces.com/problemset/problem/960/D