輸入 ：第一行輸入n（陣列內元素的個數）；第二行 n個數（陣列內的元素）；之後每行兩個數x，y（所詢問區間）。

輸出 ：分別輸出詢問區間內的最大值，最小值，和。

#include<iostream>
#include<algorithm>
#include<string>
#include<cstdio>
#include<cstring>
using namespace std;
#define inf 0x3f3f3f
#define N 200005
double t;
int n;
int a[N];
int mx[N];
int mn[N];
int sum[N];
void pushup(int rt)
{
	mx[rt] = max(mx[rt * 2], mx[rt * 2 + 1]);
	mn[rt] = min(mn[rt * 2], mn[rt * 2 + 1]);
	sum[rt] = sum[rt * 2] + sum[rt * 2 + 1];
}
void xtree(int l, int r,int rt)
{
	if (l == r)
	{
		mx[rt] = a[l];
		mn[rt] = mx[rt];
		sum[rt] = a[l];
		return;
	}
	int mid = (l + r) / 2;
	xtree(l, mid, rt * 2);
	xtree(mid + 1, r, rt * 2 + 1);
	pushup(rt);
}
//最大值
int Q1(int x, int y, int rt, int l, int r)
{
	if (x <= l && y >= r)
		return mx[rt];
	int mid = (l + r) / 2;
	int ret = 0;
	if (x <= mid) ret = max(ret, Q1(x, y, rt * 2, l, mid));
	if (y > mid)ret = max(ret, Q1(x, y, rt * 2 + 1, mid + 1, r));
	return ret;
}
//最小值
int Q2(int x, int y, int rt, int l, int r)
{
	if (x <= l &&y >= r)
		return mn[rt];
	int mid = (l + r) >> 1;
	int ret = inf;
	if (x <= mid) ret = min(ret, Q2(x, y, rt << 1, l, mid));
	if (y > mid)ret = min(ret, Q2(x, y, rt << 1 | 1, mid + 1, r));
	return ret;
}
//和
int arr1(int x, int y, int rt, int l, int r)
{
	if (x == l && y == r)
	{
		return sum[rt];
	}
	int mid = (l + r) >> 1;
	int ret = 0;
	if (y <= mid)
	{
		ret = ret + arr1(x, y, rt * 2, l, mid);
	}
	else
	{
		if (x > mid)
			ret = ret + arr1(x, y, rt * 2 + 1, mid + 1, r);
		else
		{
			ret += arr1(x, mid, rt * 2, l, mid);
			ret += arr1(mid+1, y, rt * 2 + 1, mid + 1, r);
		}
	}
	return ret;
}
int main()
{
	int x, y;
	cin >> n;
	for (int i = 1; i <= n; i++)
	{
		cin >> a[i];
	}
	xtree(1, n,1);
	while (1)
	{
		scanf("%d%d", &x, &y);
		if (x == -1)
			break;
		cout << Q1(x, y, 1, 1, n) << " " << Q2(x, y, 1, 1, n) << " " << arr1(x, y, 1, 1, n) << endl;
	}
	return 0;
}