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。

 1 /*by SilverN*/
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cstring>
 5 #include<cstdio>
 6 #include<cmath>
 7 #define lc rt<<1
 8 #define rc rt<<1|1
 9 using namespace std;
10 const int mxn=100010;
11 int read(){
12     int x=0,f=1;char ch=getchar();
13     while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
14     while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();}
15     return x*f;
16 }
17 int n,m;
18 int data[mxn];
19 struct node{
20     int mx;
21     int ml,mr;
22     int smm;
23 }t[mxn<<2],tmp0;
24 void Build(int l,int r,int rt){
25     if(l==r){t[rt].mx=t[rt].ml=t[rt].mr=data[l];t[rt].smm=data[l];return;}
26     int mid=(l+r)>>1;
27     Build(l,mid,lc);
28     Build(mid+1,r,rc);
29     t[rt].smm=t[lc].smm+t[rc].smm;
30     t[rt].mx=max(t[lc].mx,t[rc].mx);
31     t[rt].mx=max(t[lc].mr+t[rc].ml,t[rt].mx);
32     t[rt].ml=max(t[lc].ml,t[lc].smm+t[rc].ml);
33     t[rt].mr=max(t[rc].mr,t[rc].smm+t[lc].mr);
34     return;
35 }
36 node query(int L,int R,int l,int r,int rt){
37 //    printf("%d %d %d %d %d\n",L,R,l,r,rt);
38     if(L<=l && r<=R){return t[rt];}
39     int mid=(l+r)>>1;
40     node res1;
41     if(L<=mid)res1=query(L,R,l,mid,lc);
42         else res1=tmp0;
43     node res2;
44     if(R>mid)res2=query(L,R,mid+1,r,rc);
45         else res2=tmp0;
46     node res={0};
47     res.smm=res1.smm+res2.smm;
48     res.mx=max(res1.mx,res2.mx);
49     res.mx=max(res.mx,res1.mr+res2.ml);
50     res.ml=max(res1.ml,res1.smm+res2.ml);
51     res.mr=max(res2.mr,res2.smm+res1.mr);
52     return res;
53 }
54 int main(){
55     n=read();
56     int i,j,x,y;
57     for(i=1;i<=n;i++)data[i]=read();
58     Build(1,n,1);
59     m=read();
60     tmp0.ml=tmp0.mr=tmp0.mx=-1e9;tmp0.smm=0;
61     for(i=1;i<=m;i++){
62         x=read();y=read();
63         printf("%d\n",query(x,y,1,n,1).mx);
64     }
65     return 0;
66 }

 

posted @ 2016-11-30 00:07  SilverNebula  阅读(370)  评论(0)  编辑  收藏  举报
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