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+,INF=2e9+;
inline int read(){
char c=getchar();int x=,f=;
while(c<''||c>''){if(c=='-')f=-;c=getchar();}
while(c>=''&&c<=''){x=x*+c-'';c=getchar();}
return x*f;
}
int n,q,x,y;
struct node{
int sum,mx,pre,suf;
}t[N<<];
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(,,n);
q=read();
for(int i=;i<=q;i++){
x=read();y=read();
printf("%d\n",qmx(,,n,x,y));
}
}