Title Description

Sequences A[1], A[2],... A[N]. (a[i] < 15007, 1 < N < 50000). Queries are defined as follows: query (x, y)=max{a[i]+a[i+1]+... + a[j]; x < I < J < y}. Given M queries, the program must output the results of these queries.

Input and Output Format

Input format:

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.
Line M follows, with the first line containing two numbers xi and yi.
Output format:

Your program should output the results of M queries, one query per row.

Title Solution

The maximum subsection sum in the query interval is maintained by the line segment tree. lx represents the largest subsection from left to right.
rx is the largest from right to left, mx is the largest interval, and it is more convenient to use structure.

Code

#include<bits/stdc++.h>

using namespace std;
const int MAXN = 100005;

inline int rd(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0' || ch>'9') {if(ch=='-') f=-1;ch=getchar();}
    while(ch>='0' && ch<='9') {x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
    return x*f;
}

struct Node{
    int sum,rx,lx,mx;
}node[MAXN*2];

int n,m,a[MAXN];

inline void pushup(int x){
    node[x].sum=node[x<<1].sum+node[x<<1|1].sum;
    node[x].lx=max(node[x<<1].lx,node[x<<1].sum+node[x<<1|1].lx);
    node[x].rx=max(node[x<<1|1].rx,node[x<<1|1].sum+node[x<<1].rx);
    node[x].mx=max(node[x<<1].rx+node[x<<1|1].lx,
    max(node[x<<1].mx,node[x<<1|1].mx));
}

inline void build(int x,int l,int r){
    if(l==r){
        node[x].rx=node[x].lx=node[x].mx=a[l];
        node[x].sum=a[l];
        return;
    }
    int mid=(l+r)>>1;
    build(x<<1,l,mid);
    build(x<<1|1,mid+1,r);
    pushup(x);
}

inline Node query(int x,int L,int R,int l,int r){
    if(l<=L && R<=r) return node[x];
    int mid=(L+R)>>1;
    if(mid<l)  return query(x<<1|1,mid+1,R,l,r);
    if(mid>=r) return query(x<<1,L,mid,l,r);
    else{
        Node ans,a,b;
        a=query(x<<1,L,mid,l,r);b=query(x<<1|1,mid+1,R,l,r);
        ans.sum=a.sum+b.sum;
        ans.mx=max(a.mx,a.rx+b.lx),ans.mx=max(ans.mx,b.mx);
        ans.lx=max(a.lx,a.sum+b.lx);
        ans.rx=max(b.rx,b.sum+a.rx);
        return ans;
    }
}

int main(){
    n=rd();
    for(register int i=1;i<=n;i++) a[i]=rd();
    build(1,1,n);
    m=rd();
    while(m--){
        int l,r;
        l=rd();r=rd();
        printf("%d\n",query(1,1,n,l,r).mx);
    }
    return 0;
}