SPOJ GSS3 Can you answer these queries III

Time Limit: 330MS

Memory Limit: 1572864KB

64bit IO Format: %lld & %llu

Description

You are given a sequence A of N (N <= 50000) integers between -10000 and 10000. On this sequence you have to apply M (M <= 50000) operations: 
modify the i-th element in the sequence or for given x y print max{Ai + Ai+1 + .. + Aj | x<=i<=j<=y }.

Input

The first line of input contains an integer N. The following line contains N integers, representing the sequence A1..AN. 
The third line contains an integer M. The next M lines contain the operations in following form:
0 x y: modify Ax into y (|y|<=10000).
1 x y: print max{Ai + Ai+1 + .. + Aj | x<=i<=j<=y }.

Output

For each query, print an integer as the problem required.

Example

Input:
4
1 2 3 4
4
1 1 3
0 3 -3
1 2 4
1 3 3

Output:
6
4
-3

Hint

Added by:	Bin Jin
Date:	2007-08-03
Time limit:	0.330s
Source limit:	5000B
Memory limit:	1536MB
Cluster:	Cube (Intel G860)
Languages:	All except: C++ 5
Resource:	own problem

单点修改，询问区间内最大连续字段和。

@TYVJ P1427 小白逛公园

 /*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 push_up(int l,int r,int rt){
     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;
 }
 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);
     push_up(l,r,rt);
     return;
 }
 void change(int p,int v,int l,int r,int rt){
     if(l==r){
         if(p==l){t[rt].ml=t[rt].mr=t[rt].mx=t[rt].smm=v;}
         return;
     }
     int mid=(l+r)>>;
     if(p<=mid)change(p,v,l,mid,lc);
     else change(p,v,mid+,r,rc);
     push_up(l,r,rt);
     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,k;
     for(i=;i<=n;i++)data[i]=read();
     Build(,n,);
     tmp0.ml=tmp0.mr=tmp0.mx=-1e9;tmp0.smm=;
     m=read();
     for(i=;i<=m;i++){
         k=read();x=read();y=read();
         if(k){
             if(x>y)swap(x,y);
             printf("%d\n",query(x,y,,n,).mx);
         }
         else change(x,y,,n,);
     }
     return ;
 }