线段树节点信息合并

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

动态查询区间最大子段和。
维护区间和sum，最大前缀和pre，最大后缀和suf，并利用前面三个值来维护最大子段和mx

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include <queue>
#include <vector>
#include<bitset>
#include<map>
#include<deque>
using namespace std;
typedef long long LL;
const int maxn = 5e4+5;
const int mod = 77200211+233;
typedef pair<int,int> pii;
#define X first
#define Y second
#define pb push_back
//#define mp make_pair
#define ms(a,b) memset(a,b,sizeof(a))
const int inf = 0x3f3f3f3f;
#define lson l,m,2*rt
#define rson m+1,r,2*rt+1
typedef long long ll;
struct node{
    int l,r;
    int sum,mx,pre,suf;
    int mid(){
        return (l+r)>>1;
    }
}tree[maxn<<2];
void push_up(int rt){
    tree[rt].sum = tree[rt<<1].sum+tree[rt<<1|1].sum;
    tree[rt].mx = max(tree[rt<<1].suf+tree[rt<<1|1].pre,max(tree[rt<<1].mx,tree[rt<<1|1].mx));
    tree[rt].suf = max(tree[rt<<1|1].suf, tree[rt<<1|1].sum+tree[rt<<1].suf);
    tree[rt].pre = max(tree[rt<<1|1].pre+tree[rt<<1].sum, tree[rt<<1].pre);
}

void build(int l,int r,int rt){
    tree[rt].l=l,tree[rt].r=r;

    if(l==r){
        scanf("%d",&tree[rt].sum);
        tree[rt].mx=tree[rt].pre=tree[rt].suf=tree[rt].sum;
        return;
    }
    int m = tree[rt].mid();
    build(l,m,rt<<1);
    build(m+1,r,rt<<1|1);
    push_up(rt);
}
int qpre(int l,int r,int rt){
    if(l<=tree[rt].l && r>= tree[rt].r){
        return tree[rt].pre;
    }
    int m = tree[rt].mid();
    if(r<=m) return qpre(l,r,rt<<1);
    else return max(tree[rt<<1].pre, tree[rt<<1].sum+qpre(l,r,rt<<1|1));
}

int qsuf(int l,int r,int rt){
    if(l<=tree[rt].l && r>= tree[rt].r){
        return tree[rt].suf;
    }
    int m = tree[rt].mid();
    if(l>m) return qsuf(l,r,rt<<1|1);
    else return max(tree[rt<<1|1].suf, tree[rt<<1|1].sum+qsuf(l,r,rt<<1));
}
int qmx(int l,int r,int rt){
    if(l<=tree[rt].l && tree[rt].r<=r){
        return tree[rt].mx;
    }
    int m = tree[rt].mid();
    int ans = -inf;
    if(l<=m) ans = max(ans,qmx(l,r,rt<<1));
    if(r>m) ans=max(ans,qmx(l,r,rt<<1|1));
    if(l<=m&&m<r) ans = max(ans,qpre(l,r,rt<<1|1)+qsuf(l,r,rt<<1));
    return ans;
}
int main(){
    int n,q;
    scanf("%d",&n);
    build(1,n,1);
    scanf("%d",&q);
    int x,y;
    while(q--){
        scanf("%d%d",&x,&y);
        printf("%d
",qmx(x,y,1));
    }

    return 0;
}