为什么要写博客？

A - Can you answer these queries I

 

A - 你能回答这些问题吗？

 点击打开题目链接You are given a sequence A[1], A[2], ..., A. ( |A[i]| ≤ 15007 , 1 ≤ N ≤ 50000 ). A query is defined as follows: 给你一个序列A[1],A[2], ... ,A（|A[i]|≤15007,1≤N≤50000）。查询定义如下：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.给定M个查询，你的程序必须输出这些查询的结果。

Input

输入文件的第一行包含整数N.在第二行中，有N个数字.第三行包含整数M.M行跟随，行我包含2个数字xi和yi.

Output

    Your program should output the results of the M queries, one query per line.    你的程序应该输出M个查询的结果，每行一个查询。

Example

Input:
3
-1 2 3
1
1 2
Output:
2

这道题应该是线段树的算法。（先立一个flag）其中Query(x，y)= Max{a[i]+a[i+1]+ ... + a [j];x≤i≤j≤y}。意思是(我不知道)我还是去看看完全版线段树吧。。以下是别人的代码：#include <iostream>#include <cstdio>#include <algorithm>#define INF 0x3f3f3f3fusing namespace std;const int N = 50005;struct node{int sum,mx,lx,rx;inline node(int x = 0){sum = mx = lx = rx = x;}}tree[N<<2];int ql,qr;node update(node x, node y){node ans;ans.sum = x.sum + y.sum;ans.mx = max(x.rx + y.lx, max(x.mx, y.mx));ans.lx = max(x.lx, x.sum + y.lx);ans.rx = max(y.rx, y.sum + x.rx);return ans;}void build(int id,int l,int r){if(l == r){int x;scanf("%d",&x);tree[id] = node(x);return ;}int mid = (l+r)>> 1;build(id<<1,l,mid);build(id<<1|1,mid+1,r);tree[id] = update(tree[id<<1], tree[id<<1|1]);}node query(int id,int l,int r){if(ql <= l && r <= qr)return tree[id];node x(-INF), y(-INF);x.sum = y.sum = 0;int mid = (l+r)>> 1;if (ql <= mid)x = query(id<<1,l,mid);if (qr > mid)y = query(id<<1|1,mid+1,r);return update(x,y);}int main(){int n,m;while(~scanf("%d", &n)){build(1,1,n);scanf("%d",&m);while(m--){scanf("%d%d",&ql,&qr);printf("%d\n", query(1,1,n).mx);}}return 0;}我去试试看。。AC了！于是我给打了些注释。。。#include<stdio.h>const int N=50005;struct node{int sum,mx,lx,rx;}tree[N*4];int ql,qr;bool max(int a,int b){return a>b?a:b;}node update(node x,node y){//根据左子树和右子树判断本子树的值node ans;ans.sum=x.sum+y.sum;ans.mx=max(x.rx+y.lx,max(x.mx,y.mx));ans.lx=max(x.lx,x.sum+y.lx);ans.rx=max(y.rx,y.sum+x.rx);return ans;}void build(int id,int l,int r){if(l==r)//已找到该节点{int x;scanf("%d",&x);//修改该节点的值tree[id].sum=tree[id].mx=tree[id].lx=tree[id].rx=x;return ;}int mid=(l+r)/2;//[ ]|[ ]build(id*2,l,mid);//[l,mid]build(id*2+1,mid+1,r);//(mid,r]tree[id]=update(tree[id*2],tree[id*2+1]);//更新该节点的值}node query(int id,int l,int r){//查找if(ql<=l&&r<=qr)//如果当前的区间被包含return tree[id];node x,y;x.sum=x.mx4000=x.lx=x.rx=y.sum=y.mx=y.lx=y.rx=-0x3f3f3f3f;x.sum=y.sum=0;int mid=(l+r)/2;//同上if(ql<=mid)x=query(id*2,l,mid);if(qr>mid)y=query(id*2+1,mid+1,r);return update(x,y);}int main(){int n,m;while(!scanf("%d",&n)){build(1,1,n);scanf("%d",&m);while(m--){scanf("%d%d",&ql,&qr);printf("%d\n",query(1,1,n).mx);}}return 0;}