SPOJ GSS1 Can you answer these queries I

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#线段树

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本文介绍了一种使用线段树解决区间最大子段和查询问题的方法，通过构造线段树并实现查询算法，可以高效地处理大规模数据集中的查询请求。

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Description

You are given a sequence A[1], A[2], ..., A[N] . ( |A[i]| ≤ 15007 , 1 ≤ N ≤ 50000 ). A query is defined as follows:
Query(x,y) = Max { a[i]+a[i+1]+...+a[j] ; x ≤ i ≤ j ≤ y }.
Given M queries, your program must output the results of these queries.

Input

The first line of the input file contains the integer N.
In the second line, N numbers follow.
The third line contains the integer M.
M lines follow, where line i contains 2 numbers xi and yi.

Output

Your program should output the results of the M queries, one query per line.

Example

Input:
3 
-1 2 3
1
1 2
Output:
2

Input

Output

Sample Input

Sample Output

Hint

Added by:	Nguyen Dinh Tu
Date:	2006-11-01
Time limit:	0.115s-0.230s
Source limit:	5000B
Memory limit:	1536MB
Cluster:	Cube (Intel G860)
Languages:	All except: ERL JS NODEJS PERL 6 VB.net

小白逛公园的弱化版，写一下练手，第一次因为没有写return wa了。

#include<algorithm>
#include<iostream>
#include<cstdio>
using namespace std;
const int N=50005;
int n,m;
long long data[N];
struct node
{
	int l,r;
	long long sum,tot,lm,rm;
}a[4*N];
void build(int num,int l,int r)
{
	a[num].l=l,a[num].r=r;
	if(l==r)
	{
		a[num].sum=a[num].tot=a[num].lm=a[num].rm=data[l];
		return ;
	}
	int mid=(l+r)/2;
	build(2*num,l,mid);
	build(2*num+1,mid+1,r);
	a[num].sum=a[2*num].sum+a[2*num+1].sum;
	a[num].lm=max(a[2*num].lm,a[2*num].sum+a[2*num+1].lm);
	a[num].rm=max(a[2*num+1].rm,a[2*num+1].sum+a[2*num].rm);
	a[num].tot=max(max(a[2*num].tot,a[2*num+1].tot),a[2*num].rm+a[2*num+1].lm);
}
node query(int num,int l,int r)
{
	if(a[num].l>=l&&a[num].r<=r)
		return a[num];
	if(a[2*num].r>=r)
		return query(2*num,l,r);
	if(a[2*num+1].l<=l)
		return query(2*num+1,l,r);
	node u1,u2,res;
	u1=query(2*num,l,r);
	u2=query(2*num+1,l,r);
	res.sum=u1.sum+u2.sum;
	res.lm=max(u1.lm,u1.sum+u2.lm);
	res.rm=max(u2.rm,u2.sum+u1.rm);
	res.tot=max(max(u1.tot,u2.tot),u1.rm+u2.lm);
	return res;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%lld",&data[i]);
	build(1,1,n);
	scanf("%d",&m);
	int x,y;
	while(m--)
	{
		scanf("%d%d",&x,&y);
		printf("%lld\n",query(1,x,y).tot);
	}
	return 0;
}