CodeForces 696A：Lorenzo Von Matterhorn（map的用法）

http://codeforces.com/contest/697/problem/C

C. 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 ≤ 10¹⁸, v ≠ u, 1 ≤ w ≤ 10⁹ 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

Note

In the example testcase:

Here are the intersections used:

1. Intersections on the path are 3, 1, 2 and 4.
2. Intersections on the path are 4, 2 and 1.
3. Intersections on the path are only 3 and 6.
4. Intersections on the path are 4, 2, 1 and 3. Passing fee of roads on the path are 32, 32 and 30 in order. So answer equals to32 + 32 + 30 = 94.
5. Intersections on the path are 6, 3 and 1.
6. Intersections on the path are 3 and 7. Passing fee of the road between them is 0.
7. Intersections on the path are 2 and 4. Passing fee of the road between them is 32 (increased by 30 in the first event and by 2 in the second).

题意：在一个类似于树的图中，求从一个节点到另一个节点需要走过的距离，初始为0，输入为1时对u到v进行更新，输入为2时求u到v的距离。

思路：用map放每一个节点的对应边的权值，然后进行更新或者查询

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <iostream>
 #include <map>
 using namespace std;
 #define N 10005
 map<long long,long long> mp;

 void add(long long a,long long w)
 {
     if(mp.find(a)==mp.end()) mp[a]=;
     mp[a]+=w;
 }

 long long val(long long a)
 {
     if(mp.find(a)==mp.end()) return ;
     return mp[a];
 }

 int main()
 {
     int t;
     cin>>t;
     while(t--){
         int s;
         cin>>s;
         if(s==){
             long long u,v,w;
             cin>>u>>v>>w;
             while(u!=v){
                 //对节点进行更新，直到更新到根节点
                 if(u>v){
                     add(u,w);
                     u>>=;
                 }
                 else{
                     add(v,w);
                     v>>=;
                 }
             }
         }
         else{
             long long u,v,ans=;
             cin>>u>>v;
             while(v!=u){
                 //查询
                 if(u>v){
                     ans+=val(u);
                     u>>=;
                 }
                 else{
                     ans+=val(v);
                     v>>=;
                 }
             }
             cout<<ans<<endl;
         }
     }
     return ;
 }