recursion有一个整数序列a[n]。现在recursion有m次询问，每次她想知道Max { A[i]+A[i+1]+...+A[j] ; x1 <= i <= y1 , x2 <= j <= y2 , x1 <= x2 , y1 <= y2 }。这么简单的题，recursion当然会做啦，但是为了维持她的傲娇属性，她决定考考你。
Input
输入的第一行为数据组数。对于每组数据，第一行包含一个正整数n和长度为n的序列a[n]。接下来一行有一个正整数m。下面m行分别描述m个询问，每行包含四个整数x1,y1,x2,y2。
Output
对于每组数据输出m行，分别表示m个询问的答案
Sample Input
2
6 3 -2 1 -4 5 2
2
1 1 2 3
1 3 2 5
1 1
1
1 1 1 1
Sample Output
2
3
1
Hint
|A[i]|<=10000,1<=N<=10000,1<=M<=10000
思路：
首先用https://vjudge.net/problem/SPOJ-GSS3 这题的模板可以维护正常的区间询问，单点修改的线段树维护区间最大子段和问题。
然后对于题目的给定两个区间中分别选一个l和r问题，
我们进行分类讨论，我们看答案可能是哪种情况。
首先，如果x2>y1 那么答案就是 ( x1~y1 )中最大后缀数值+ ( y1~x2 ) 区间的数值sum和+ ( x2 + y2 中区间的最大前缀和 )
否则 两个区间就一定有交叉的分布
即 数值的从小到大的顺序是这样： x1，x2，y1 , y2
那么答案可能是以下三种情况：
1、x2~y1 中的最大子段和。
2、l在 x1~x2 之间，y在x2~y2区间
3、l在x1~y1 之间，y在y1~y2 区间，
这三种情况就包含了所有可能。
细节见代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), ' ', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2) { ans = ans * a % MOD; } a = a * a % MOD; b /= 2;} return ans;}
inline void getInt(int *p);
const int maxn = 50000 + 7;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
struct node {
int l;
int r;
ll num;
ll lm;
ll sum;
ll rm;
} segment_tree[maxn << 2];
int n;
void pushup(int rt)
{
segment_tree[rt].sum = segment_tree[rt << 1].sum + segment_tree[rt << 1 | 1].sum;
segment_tree[rt].lm = max(segment_tree[rt << 1].lm, segment_tree[rt << 1].sum + segment_tree[rt << 1 | 1].lm);
segment_tree[rt].rm = max(segment_tree[rt << 1 | 1].rm, segment_tree[rt << 1 | 1].sum + segment_tree[rt << 1].rm);
segment_tree[rt].num = max(segment_tree[rt << 1].num, segment_tree[rt << 1 | 1].num);
segment_tree[rt].num = max(segment_tree[rt].num, segment_tree[rt << 1].rm + segment_tree[rt << 1 | 1].lm);
}
void build(int rt, int l, int r)
{
segment_tree[rt].l = l;
segment_tree[rt].r = r;
if (l == r) {
scanf("%lld", &segment_tree[rt].num);
segment_tree[rt].lm = segment_tree[rt].rm = segment_tree[rt].num;
segment_tree[rt].sum = segment_tree[rt].num;
return ;
}
int mid = (l + r) >> 1;
build(rt << 1, l, mid);
build(rt << 1 | 1, mid + 1, r);
pushup(rt);
}
node ask(int rt, int l, int r)
{
if (l > r) {
return node{0, 0, 0, 0, 0, 0};
}
if (segment_tree[rt].l == l && segment_tree[rt].r == r) {
return segment_tree[rt];
}
int mid = (segment_tree[rt].r + segment_tree[rt].l) >> 1;
if (l > mid) {
return ask(rt << 1 | 1, l, r);
} else if (r <= mid) {
return ask(rt << 1, l, r);
} else {
node res1 = ask(rt << 1, l, mid);
node res2 = ask(rt << 1 | 1, mid + 1, r);
node res;
res.sum = res1.sum + res2.sum;
res.lm = max(res1.lm, res1.sum + res2.lm);
res.rm = max(res2.rm, res2.sum + res1.rm);
res.num = max(res1.num, res2.num);
res.num = max(res.num, res1.rm + res2.lm);
return res;
}
}
void update(int rt, int x, int val)
{
if (segment_tree[rt].l == x && segment_tree[rt].r == x) {
segment_tree[rt].num = val;
segment_tree[rt].lm = val;
segment_tree[rt].rm = val;
segment_tree[rt].sum = val;
return ;
}
int mid = (segment_tree[rt].l + segment_tree[rt].r) >> 1;
if (x <= mid) {
update(rt << 1, x, val);
} else {
update(rt << 1 | 1, x, val);
}
pushup(rt);
}
ll solve(int x1, int y1, int x2, int y2)
{
if (y1 < x2) {
ll res = ask(1, x1, y1).rm;
res += ask(1, y1 + 1, x2 - 1).sum;
res += ask(1, x2, y2).lm;
return res;
} else {
ll res = ask(1, x2, y1).num;
res = max(res, ask(1, x1, x2).rm + ask(1, x2, y2).lm - ask(1, x2, x2).sum);
res = max(res, ask(1, x1, y1).rm + ask(1, y1, y2).lm - ask(1, y1, y1).sum);
return res;
}
}
int main()
{
//freopen("D:\code\text\input.txt","r",stdin);
//freopen("D:\code\text\output.txt","w",stdout);
int t;
scanf("%d", &t);
while (t--) {
scanf("%d", &n);
build(1, 1, n);
int m;
scanf("%d", &m);
while (m--) {
int x1, x2, y1, y2;
scanf("%d %d %d %d", &x1, &y1, &x2, &y2);
printf("%lld
", solve(x1, y1, x2, y2));
}
}
return 0;
}
inline void getInt(int *p)
{
char ch;
do {
ch = getchar();
} while (ch == ' ' || ch == '
');
if (ch == '-') {
*p = -(getchar() - '0');
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 - ch + '0';
}
} else {
*p = ch - '0';
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 + ch - '0';
}
}
}