HDU 5662 YJQQQAQ and the function

Problem Description
YJQQQAQ has an array 
A  of length 
n . He defines a function 
fl,r,k  where 
l,r,k  are positive integers that satisfies 
l≤r  and 
r×k≤n , and the value of the function equals to 
p×q×⌊k√⌋  where 
p  equals to the sum value of 
Al×k,A(l+1)×k,...,Ar×k  and 
q  equals to the minimal value of them. YJQQQAQ wants to choose the positive integers 
l,r,k  carefully to maximize the value of the function.
 
Input
The first line contains an integer 
T(1≤T≤3) ——The number of the test cases. For each test case:
The first line contains an integers 
n(1≤n≤300,000) .
The second line contains 
n  integers describing the given array 
A , the 
i th integer is 
Ai(1≤Ai≤1,000,000) . Between each two adjacent integers there is a white space separated.
 
Output
For each test case, the only line contains the only integer that is the maximum value of the function.
 
Sample Input

   
   
   
   

    
    
    
    1
3
2 3 1
   
   
   
   

 
Sample Output

   
   
   
   

    
    
    
    10


    
    
    
    
 
     
 
      Hint 
     
 When and only when $l=1,r=2,k=1$, the value of the function is the maximum. 
    
    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
    
#pragma comment(linker, "/STACK:1024000000,1024000000") 
#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<bitset>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<functional>
using namespace std;
typedef __int64 LL;
const int low(int x) { return x&-x; }
const int INF = 0x7FFFFFFF;
const int mod = 1e9 + 7;
const int maxn = 3e5 + 10;
int T, n, m, a[maxn], b[maxn], l[maxn], r[maxn];
LL sum[maxn];

int main() {
	scanf("%d", &T);
	while (T--)
	{
		scanf("%d", &n);
		for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
		LL ans = 0;
		for (int i = 1; i <= n; i++)
		{
			sum[0] = m = 0;
			for (int j = 1; j*i <= n; j++)
			{
				b[++m] = a[j*i];
				sum[m] = sum[m - 1] + b[m];
			}
			stack<int> p;
			for (int j = 1; j <= m; j++)
			{
				while (!p.empty() && b[p.top()] > b[j]) r[p.top()] = j - 1, p.pop();
				p.push(j);
			}
			while (!p.empty()) r[p.top()] = m, p.pop();
			for (int j = m; j; j--)
			{
				while (!p.empty() && b[p.top()] > b[j]) l[p.top()] = j + 1, p.pop();
				p.push(j);
			}
			while (!p.empty()) r[p.top()] = 1, p.pop();
			for (int j = 1; j <= m; j++) ans = max(ans, (sum[r[j]] - sum[l[j] - 1])*b[j] * (int)sqrt(i));
		}
		printf("%I64d
", ans);
	}
	return 0;
}