南昌 Max answer

Posted 2021-11-03 -citywall123

https://nanti.jisuanke.com/t/38228

Alice has a magic array. She suggests that the value of a interval is equal to the sum of the values in the interval, multiplied by the smallest value in the interval.

Now she is planning to find the max value of the intervals in her array. Can you help her?

Input

First line contains an integer n(1≤n≤5×105n(1 \le n \le 5 \times 10 ^5n(1≤n≤5×105).

Second line contains nnn integers represent the array a(−105≤ai≤105)a (-10^5 \le a_i \le 10^5)a(−105≤ai?≤105).

Output

One line contains an integer represent the answer of the array.

样例输入

5
1 2 3 4 5

样例输出

36
题意：求区间和乘以区间内最小的一个数的最大值，包含正负

题解：前缀和预处理，单调栈找到每一个数向左延伸的最大距离和向右延伸的最大距离，分别记录这两个坐标。
求最大值的时候分正负处理：
正数：mx=max(mx,a[i]*(sum[r[i]]-sum[l[i]]+a[l[i]];
负数：暴力加和，[l[i],r[i]]区间内的数都加起来，每次更新最小和（因为是负数，乘以a[i]后就是最大值）

#include<iostream>
#include<string.h>
#include<stack>
#define ll long long
using namespace std;
ll l[500005], r[500005], a[500005], sum[500005];
ll n,mx;
int main()
{
    while (~scanf("%lld", &n))
    {
        for (int i = 0; i < n; i++)//刚开始是从1开始输入的，然后前缀和也是从1开始处理的，一位可以简单一点，但是因为
        {                          //栈是从0开始处理的，下面的a[p.top()]可能为a[0],导致一直TLE，找这个错误到自闭。。。
            scanf("%lld", &a[i]);
        }
        sum[0] = a[0];
        for (int i = 1; i < n; i++)
        {
            sum[i] = sum[i - 1] + a[i];
        }
        stack<ll>p;
        for (int i = 0; i < n; i++)
        {
            while (!p.empty() && a[p.top()] >= a[i])//按照大于等于a[i]的规则，从a[i]开始，向左边可以延伸最远的数的下标
                p.pop();
            if (p.empty())//说明a[i]左边的所有数都大于等于a[i]
                l[i] = 0;
            else//找到就记录从a[i]开始向左边寻找大于等于a[i]的最大边界下标
                l[i] = p.top() + 1;
            p.push(i);
        }
        while (!p.empty())
            p.pop();
        for (int i = n - 1; i >= 0; i--)//往右边找第一个比a[i]小的数
        {
            while (!p.empty() && a[p.top()] >= a[i])
                p.pop();
            if (p.empty())
                r[i] = n - 1;
            else
                r[i] = p.top() - 1;
            p.push(i);
        }
        mx = 0;
        for (int i = 0; i < n; i++)
        {
            if (a[i] >= 0)
                mx = max(mx, a[i] * (sum[r[i]] - sum[l[i]] + a[l[i]]));//这里是a[l[i]]
            else
            {
                ll ans = 0, m1 = 0, m2 = 0;
                for (int j = i + 1; j <= r[i] && j < n; j++)
                {
                    ans = ans + a[j];
                    m1 = min(m1, ans);
                }
                ans = 0;
                for (int j = i - 1; j >= l[i] && j >= 0; j--)
                {
                    ans = ans + a[j];
                    m2 = min(ans, m2);
                }
                mx = max(mx, a[i] * (m1 + m2 + a[i]));
            }
        }

        printf("%lld\n", mx);
        // for(int i=1;i<=n;i++)
        //     cout<<l[i]<<‘ ‘;
        // cout<<endl;
        // for(int i=1;i<=n;i++)
        //     cout<<r[i]<<‘ ‘;
        // cout<<endl;

    }
    return 0;
}