SPOJ GSS1 Can you answer these queries I

Time Limit: 115MS

Memory Limit: 1572864KB

64bit IO Format: %lld & %llu

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

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

询问区间内和最大的连续子序列。没有修改操作。

线段树维护即可。

 /*by SilverN*/
 #include<algorithm>
 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<cmath>
 #define lc rt<<1
 #define rc rt<<1|1
 using namespace std;
 const int mxn=;
 int read(){
     int x=,f=;char ch=getchar();
     while(ch<'' || ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>='' && ch<=''){x=x*+ch-'';ch=getchar();}
     return x*f;
 }
 int n,m;
 int data[mxn];
 struct node{
     int mx;
     int ml,mr;
     int smm;
 }t[mxn<<],tmp0;
 void Build(int l,int r,int rt){
     if(l==r){t[rt].mx=t[rt].ml=t[rt].mr=data[l];t[rt].smm=data[l];return;}
     int mid=(l+r)>>;
     Build(l,mid,lc);
     Build(mid+,r,rc);
     t[rt].smm=t[lc].smm+t[rc].smm;
     t[rt].mx=max(t[lc].mx,t[rc].mx);
     t[rt].mx=max(t[lc].mr+t[rc].ml,t[rt].mx);
     t[rt].ml=max(t[lc].ml,t[lc].smm+t[rc].ml);
     t[rt].mr=max(t[rc].mr,t[rc].smm+t[lc].mr);
     return;
 }
 node query(int L,int R,int l,int r,int rt){
 //    printf("%d %d %d %d %d\n",L,R,l,r,rt);
     if(L<=l && r<=R){return t[rt];}
     int mid=(l+r)>>;
     node res1;
     if(L<=mid)res1=query(L,R,l,mid,lc);
         else res1=tmp0;
     node res2;
     if(R>mid)res2=query(L,R,mid+,r,rc);
         else res2=tmp0;
     node res={};
     res.smm=res1.smm+res2.smm;
     res.mx=max(res1.mx,res2.mx);
     res.mx=max(res.mx,res1.mr+res2.ml);
     res.ml=max(res1.ml,res1.smm+res2.ml);
     res.mr=max(res2.mr,res2.smm+res1.mr);
     return res;
 }
 int main(){
     n=read();
     int i,j,x,y;
     for(i=;i<=n;i++)data[i]=read();
     Build(,n,);
     m=read();
     tmp0.ml=tmp0.mr=tmp0.mx=-1e9;tmp0.smm=;
     for(i=;i<=m;i++){
         x=read();y=read();
         printf("%d\n",query(x,y,,n,).mx);
     }
     return ;
 }