Neat Tree

可以分开计算，先计算所有区间内的最大值之和，再减去所有区间的最小值之和。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <time.h>
#include <string>
#include <stack>
#include <algorithm>
using namespace std;
#define gc p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1000000, stdin), p1 == p2) ? EOF : *p1++;
inline int read(){ static char buf[1000000], *p1 = buf, *p2 = buf; register int x = false; register char ch = gc; register bool sgn = false; while (ch != '-' && (ch < '0' || ch > '9')) ch = gc; if (ch == '-') sgn = true, ch = gc; while (ch >= '0'&& ch <= '9') x = (x << 1) + (x << 3) + (ch ^ 48), ch = gc; return sgn ? -x : x; }
#ifdef LOCAL
#define TIME cout << "RuningTime: " << clock() << "ms\n", 0
#else
#define TIME 0
#endif
typedef long long ll;
const int N = 3e6 + 10;
const int mod = 998244353;
int a[N];
ll f[N];
int stk[N];
int n;
ll calc1(int flag)
{
    for (int i = 1; i <= n; i++)
        f[i] = 0;
    int top = 0;
    ll ans = 0, sum = 0;
    for (int i = 1; i <= n; i++)
    {
        while (top && (flag == 0 ? a[stk[top]] > a[i] : a[stk[top]] < a[i])) //递增 or 递减
            top--;
        f[i] += 1LL * (i - stk[top]) * a[i];
        f[i] += f[stk[top]];
        stk[++top] = i;
        ans += f[i];
    }
    return ans;
}
int main()
{
#ifdef LOCAL
    freopen("E:/input.txt", "r", stdin);
#endif
    while (cin >> n)
    {
        for (int i = 1; i <= n; i++)
            scanf("%d", &a[i]);
        cout << calc1(1) - calc1(0) << endl;
    }
    return TIME;
}