SPOJ GSS1 Can you answer these queries I[线段树]

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

---

动态最大子段和
维护区间和sum，最大连续和mx，最大前缀和pre，最大后缀和suf
用了三个查询，子段，前缀和后缀
考虑最大的起点和终点位置然后更新

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#define m ((l+r)>>1)
#define lson o<<1,l,m
#define rson o<<1|1,m+1,r
#define lc o<<1
#define rc o<<1|1
using namespace std;
typedef long long ll;
const int N=5e5+5,INF=2e9+5;
inline int read(){
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}
int n,q,x,y;
struct node{
    int sum,mx,pre,suf;
}t[N<<2];
void merge(int o){
    t[o].sum=t[lc].sum+t[rc].sum;
    t[o].mx=max(t[lc].suf+t[rc].pre,max(t[lc].mx,t[rc].mx));
    t[o].pre=max(t[lc].pre,t[lc].sum+t[rc].pre);
    t[o].suf=max(t[rc].suf,t[rc].sum+t[lc].suf);
}
void build(int o,int l,int r){
    if(l==r) t[o].sum=t[o].mx=t[o].pre=t[o].suf=read();
    else{
        build(lson);
        build(rson);
        merge(o);
    }
}
int qpre(int o,int l,int r,int ql,int qr){
    if(ql<=l&&r<=qr) return t[o].pre;
    else if(qr<=m) return qpre(lson,ql,qr);
    else return max(qpre(lson,ql,qr),t[lc].sum+qpre(rson,ql,qr));
}
int qsuf(int o,int l,int r,int ql,int qr){
    if(ql<=l&&r<=qr) return t[o].suf;
    else if(m<ql) return qsuf(rson,ql,qr);
    else return max(t[rc].suf,t[rc].sum+qsuf(lson,ql,qr));
}
int qmx(int o,int l,int r,int ql,int qr){
    if(ql<=l&&r<=qr) return t[o].mx;
    else{
        int ans=-INF;
        if(ql<=m) ans=max(ans,qmx(lson,ql,qr));
        if(m<qr) ans=max(ans,qmx(rson,ql,qr));
        if(ql<=m&&m<qr) ans=max(ans,qsuf(lson,ql,qr)+qpre(rson,ql,qr));
        return ans;
    }
}

int main(){
    n=read();
    build(1,1,n);
    q=read();
    for(int i=1;i<=q;i++){
        x=read();y=read();
        printf("%d
",qmx(1,1,n,x,y));
    }
}