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

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 ton, in the order from the highest to the lowest, the volume of thei-th vessel is a_i liters.

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

Initially, all the vessels are empty. In some vessels water is poured. All the water that overflows from thei-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:

Add x_i liters of water to thep_i-th vessel;
Print the number of liters of water in the k_i-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 containsn integers a₁, a₂, ..., a_n — the vessels' capacities (1 ≤ a_i ≤ 10⁹). The vessels' capacities do not necessarily increase from the top vessels to the bottom ones (see the second sample). The third line contains integerm — the number of queries (1 ≤ m ≤ 2·10⁵). Each of the nextm lines contains the description of one query. The query of the first type is represented as "1p_ix_i", the query of the second type is represented as "2k_i" (1 ≤ p_i ≤ n,1 ≤ x_i ≤ 10⁹,1 ≤ k_i ≤ n).

Output

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

Sample test(s)

Input

Output

4
5
8

Input

Output

7
10
5

题意：给一个从上到下的漏斗，当漏斗一满时，就会溢到下一个漏斗（就算下一个小过上面的一个也一样漏到下一个），然后他给你两种状态，一种是注水，一种是查询

其实查询很简单，就是维护一个数组代表着n个漏斗的水量。然后直接查询就行了。最重要的是如何最多在nlog(n)的条件下可以维护数组。刚开始我写的很暴力，n^2,超时了

其实这道题就是人工模拟链表。用数组来维护链表的指针。然后更新。。

反省一下：下次要不断地思考问题。因为一道acm的题目不仅要会算法还要懂得分析这道题的解决思路，这才是王道。还有自己做题速度实在是太慢了。。

附我ac代码

#include<cstdio>
#include<cstring>
int nv[200009],v[200009];
int f[200009],go[200009];
int main()
{
    int n,i,j,k,vv,m;
    while(~scanf("%d",&n))
    {
        memset(nv,0,sizeof(nv));
        for(i=0;i<n;i++)
        {
            scanf("%d",&v[i]);
            go[i] = i+1;
        }
        scanf("%d",&m);
        int select;
        for(i=0;i<m;i++)
        {
            scanf("%d",&select);
            if(select==2)
            {
                scanf("%d",&k);
                printf("%d\n",nv[k-1]);
            }
            else
            {
                scanf("%d%d",&k,&vv);
                k--;
                int kk = k;
                int r = 0;
            
                while(vv>0 && kk<n)
                {
                    
                    if(vv<=v[kk]-nv[kk])
                    {
                        nv[kk] += vv;
                        vv = 0;
                    }
                    else
                    {
                        vv -= (v[kk]-nv[kk]);
                        nv[kk] = v[kk];
                        f[r++] = k;
                        kk = go[kk];
                        
                    }
                    
                }
                for(j=0;j<r;j++)
                    go[f[j]] = kk; 
            }
        }
    }
    return 0;

}

TLE代码

#include<cstdio>
#include<cstring>
int nv[200009],v[200009];
int f[200009];
int main()
{
	int n,i,j,k,vv,m;
	while(~scanf("%d",&n))
	{
		memset(f,0,sizeof(f));
		for(i=1;i<=n;i++)
		{
			scanf("%d",&v[i]);
			nv[i] = v[i];
		}
		scanf("%d",&m);
		int select;
		for(i=0;i<m;i++)
		{
			scanf("%d",&select);
			if(select==2)
			{
				scanf("%d",&k);
				printf("%d\n",nv[k]-v[k]);
			}
			else
			{
				scanf("%d%d",&k,&vv);
				while(vv>0 && k<=n)
				{
					if(!f[k])
					{
						if(vv>=v[k])
						{
							vv-=v[k];
							v[k] = 0;
							f[k] = 1;
						}
						else
						{
							v[k]-=vv;
							vv = 0;
						}
					}
					k++;
				}
			}
		}
	}
	return 0;

}

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

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