Codeforces Round #218 (Div. 2) D. Vessels

D. Vessels

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

There is a system of n vessels arranged one above the other as shown in the figure below. Assume that the vessels are numbered from 1 to n, in the order from the highest to the lowest, the volume of the i-th vessel is ailiters.

Initially, all the vessels are empty. In some vessels water is poured. All the water that overflows from the i-th vessel goes to the (i + 1)-th one. The liquid that overflows from the n-th vessel spills on the floor.

Your task is to simulate pouring water into the vessels. To do this, you will need to handle two types of queries:

1. Add xi liters of water to the pi-th vessel;
2. Print the number of liters of water in the ki-th vessel.

When you reply to the second request you can assume that all the water poured up to this point, has already overflown between the vessels.

Input

The first line contains integer n — the number of vessels (1 ≤ n ≤ 2·105). The second line contains n integersa1, a2, ..., an — the vessels' capacities (1 ≤ ai ≤ 109). The vessels' capacities do not necessarily increase from the top vessels to the bottom ones (see the second sample). The third line contains integer m — the number of queries (1 ≤ m ≤ 2·105). Each of the next m lines contains the description of one query. The query of the first type is represented as "1 pi xi", the query of the second type is represented as "2 ki" (1 ≤ pi ≤ n, 1 ≤ xi ≤ 109, 1 ≤ ki ≤ n).

Output

For each query, print on a single line the number of liters of water in the corresponding vessel.

Sample test(s)

input

2
5 10
6
1 1 4
2 1
1 2 5
1 1 4
2 1
2 2

output

4
5
8

input

3
5 10 8
6
1 1 12
2 2
1 1 6
1 3 2
2 2
2 3

output

7
10
5

这个题的做法应该有很多。

这种写法有点像并查集。

其实也就是模拟的思想，把已经满了的节点delete，不断的调整后继节点。每个点大概访问2次左右，跑了140ms。

#include <iostream>
#include <string>
#include <string.h>
#include <map>
#include <stdio.h>
#include <algorithm>
#include <queue>
#include <vector>
#include <math.h>
#include <set>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std ;
typedef long long LL ;
const int Max_N =  ;
int  Right[Max_N] ;
int  now[Max_N] ;
int  limit[Max_N] ;
int  N ,M ;

int find_right(int id,int x){
     if(id != -){
        int leave = limit[id] - now[id] ;
        if(leave >= x){
            now[id] += x ;
            return id ;
        }
        else{
            now[id] = limit[id] ;
            return Right[id] = find_right(Right[id],x - leave) ;
        }
     }
     else
        return - ;
}

int main(){
    int kind ,id ,x ;
    while(scanf("%d",&N)!=EOF){
         for(int i =  ; i <= N ; i++){
             scanf("%d",&limit[i]) ;
             now[i] =  ;
             Right[i] = i+ ;
         }
         Right[N] = - ;
         scanf("%d",&M) ;
         while(M--){
               scanf("%d%d",&kind,&id) ;
               if(kind == ){
                  scanf("%d",&x) ;
                  find_right(id,x) ;
               }
               else
                  printf("%d\n",now[id]) ;
         }
    }
    return  ;
}