codeforces 696A A. Lorenzo Von Matterhorn(水题)

题目链接:

A. Lorenzo Von Matterhorn

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Barney lives in NYC. NYC has infinite number of intersections numbered with positive integers starting from 1. There exists a bidirectional road between intersections i and 2i and another road between i and 2i + 1 for every positive integer i. You can clearly see that there exists a unique shortest path between any two intersections.

Initially anyone can pass any road for free. But since SlapsGiving is ahead of us, there will q consecutive events happen soon. There are two types of events:

1. Government makes a new rule. A rule can be denoted by integers v, u and w. As the result of this action, the passing fee of all roads on the shortest path from u to v increases by w dollars.

2. Barney starts moving from some intersection v and goes to intersection u where there's a girl he wants to cuddle (using his fake name Lorenzo Von Matterhorn). He always uses the shortest path (visiting minimum number of intersections or roads) between two intersections.

Government needs your calculations. For each time Barney goes to cuddle a girl, you need to tell the government how much money he should pay (sum of passing fee of all roads he passes).

Input

The first line of input contains a single integer q (1 ≤ q ≤ 1 000).

The next q lines contain the information about the events in chronological order. Each event is described in form 1 v u w if it's an event when government makes a new rule about increasing the passing fee of all roads on the shortest path from u to v by w dollars, or in form 2 v u if it's an event when Barnie goes to cuddle from the intersection v to the intersection u.

1 ≤ v, u ≤ 10^18, v ≠ u, 1 ≤ w ≤ 10^9 states for every description line.

Output

For each event of second type print the sum of passing fee of all roads Barney passes in this event, in one line. Print the answers in chronological order of corresponding events.

Example

input

7
1 3 4 30
1 4 1 2
1 3 6 8
2 4 3
1 6 1 40
2 3 7
2 2 4

output

94
0
32

题意：

 

i与2*i和2*i+1相连的一个图,两种操作,一种是在u,v的道路上的每一条边权值加w,一种是询问u,v之间道路上的边权值和;

 

思路：

 

可以直接暴力,可以找出最近功公共祖先,然后找这两个点与公共祖先之间把边的权值加上;

 

AC代码：

#include <bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<''||CH>'';F= CH=='-',CH=getchar());
    for(num=;CH>=''&&CH<='';num=num*+CH-'',CH=getchar());
    F && (num=-num);
}
int stk[], tp;
template<class T> inline void print(T p) {
    if(!p) { puts(""); return; }
    while(p) stk[++ tp] = p%, p/=;
    while(tp) putchar(stk[tp--] + '');
    putchar('\n');
}

const LL mod=1e9+;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=3e6+;
const int maxn=3e6;
const double eps=1e-;

int q;
map<LL,LL>mp,vis;
LL getfa(LL x,LL y)
{
    vis.clear();
    while(x)vis[x]=,x>>=;
    while()
    {
        if(vis[y])return y;
        y>>=;
    }
}
int main()
{
        read(q);
        while (q--)
        {
            int f;
            LL u,v,w;
            read(f);
            if(f==)
            {
                read(u);read(v);read(w);
                LL fa=getfa(u,v);
                while(u!=fa)
                {
                    mp[u]=mp[u]+w;
                    u>>=;
                }
                while(v!=fa)
                {
                    mp[v]=mp[v]+w;
                    v>>=;
                }
            }
            else
            {
                read(u);read(v);
                LL fa=getfa(u,v),ans=;
                while(u!=fa)
                {
                    ans+=mp[u];
                    u>>=;
                }
                while(v!=fa)
                {
                    ans+=mp[v];
                    v>>=;
                }
                print(ans);
            }
        }
        return ;
}