线段树 区间合并

sum表示节点的和

max表示节点的最大值

lsum表示节点从左边开始的最大值

rsum表示节点从右边开始的最大值

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 2 311 2 Output: 2

#include <stdio.h>#include <stdlib.h>#include <malloc.h>#include <limits.h>#include <string.h>#include <string>#include <queue>#include <vector>#include <set>#include <map>#include <deque>#include <map>#include <algorithm>#include <iostream>using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define MAXN 500000 + 100int sum[MAXN<<2],lsum[MAXN<<2],rsum[MAXN<<2],maxx[MAXN<<2];int num[MAXN];int Max(int a,int b){    return a>b?a:b;}void PushUp(int rt){    sum[rt] = sum[rt<<1] + sum[rt<<1|1];    lsum[rt] = Max(lsum[rt<<1],sum[rt<<1]+lsum[rt<<1|1]);    rsum[rt] = Max(rsum[rt<<1|1],sum[rt<<1|1]+rsum[rt<<1]);    maxx[rt] = Max(maxx[rt<<1],maxx[rt<<1|1]);    maxx[rt] = Max(maxx[rt],lsum[rt]);    maxx[rt] = Max(maxx[rt],rsum[rt]);    maxx[rt] = Max(maxx[rt],rsum[rt<<1]+lsum[rt<<1|1]);}void build(int l,int r,int rt){    if(l == r){        sum[rt] = lsum[rt] = rsum[rt] = maxx[rt] = num[l];        return ;    }    int m = (l+r)>>1;    build(lson);    build(rson);    PushUp(rt);}int query_l(int L,int R,int l,int r,int rt){    if(L==l && r==R){        return rsum[rt];    }    int m = (l+r)>>1;    if(L > m){        return query_l(L,R,rson);    }    else{        int s = rsum[rt<<1|1];        int ls = query_l(L,m,lson)+sum[rt<<1|1];//sum[rt<<1+1]+rsum[rt<<1];        return Max(s,ls);    }}int query_r(int L,int R,int l,int r,int rt){    if(L==l && r==R){        return lsum[rt];    }    int m = (l+r)>>1;    if(R <= m){        return query_r(L,R,lson);    }    else{        int s = lsum[rt<<1];        int ls = sum[rt<<1]+query_r(m+1,R,rson);//sum[rt<<1]+lsum[rt<<1|1]        return Max(s,ls);    }}int query(int L,int R,int l,int r,int rt){    if(L==l && r==R){        return maxx[rt];    }    int m = (l+r)>>1;    if(m >= R){//等于号位置变了下都超时。。。。。。(这样查询的时间可能是最快的把)        return query(L,R,lson);    }    else if(L > m){        return query(L,R,rson);    }    else{        int ls = query(L,m,lson);        int rs = query(m+1,R,rson);        int ms = query_l(L,m,lson)+query_r(m+1,R,rson);//rsum[rt<<1]+lsum[rt<<1|1];        return Max(Max(ls,rs),ms);    }}int main(){    int n,m,a,b;    int i;    while(~scanf("%d",&n)){        for(i=1;i<=n;i++){            scanf("%d",&num[i]);        }        build(1,n,1);        scanf("%d",&m);        while(m--){            scanf("%d%d",&a,&b);            printf("%d\n",query(a,b,1,n,1));        }    }    return 0;}