CodeForces - 660F：Bear and Bowling 4（DP+斜率优化）

 

Limak is an old brown bear. He often goes bowling with his friends. Today he feels really good and tries to beat his own record!

For rolling a ball one gets a score — an integer (maybe negative) number of points. Score for the i-th roll is multiplied by i and scores are summed up. So, for k rolls with scores s₁, s₂, ..., s_k, the total score is . The total score is 0 if there were no rolls.

Limak made n rolls and got score a_i for the i-th of them. He wants to maximize his total score and he came up with an interesting idea. He can say that some first rolls were only a warm-up, and that he wasn't focused during the last rolls. More formally, he can cancel any prefix and any suffix of the sequence a₁, a₂, ..., a_n. It is allowed to cancel all rolls, or to cancel none of them.

The total score is calculated as if there were only non-canceled rolls. So, the first non-canceled roll has score multiplied by 1, the second one has score multiplied by 2, and so on, till the last non-canceled roll.

What maximum total score can Limak get?

Input

The first line contains a single integer n (1 ≤ n ≤ 2·10⁵) — the total number of rolls made by Limak.

The second line contains n integers a₁, a₂, ..., a_n (|a_i| ≤ 10⁷) — scores for Limak's rolls.

Output

Print the maximum possible total score after cancelling rolls.

Examples

Input

6
5 -1000 1 -3 7 -8

Output

16

Input

5
1000 1000 1001 1000 1000

Output

15003

Input

3
-60 -70 -80

Output

0

题意：给定一段数列，现在叫你取其中一段，第一位*1，第二位*2...求最大。

思路：设前缀和和是sum[n]=a[1]+a[2]+...a[n]，原先的dp[n]=a[1]*1+a[2]*2+a[3]*3+..a[n]*n;

           那么现在的max[i]=max:dp[i]-dp[j]-(sum[i]-sum[j])*j; =-sum[i]*j+j*sum[j]-dp[j]+dp[j] 

           显然可以斜率优化，由于K=sum[i]并不是单调递增的，所以我们维护凸包不能弹出队首，而需要二分。

和BZOJ2726一样，就补多说了。

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=200010;
ll s[maxn],dp[maxn],q[maxn],top,ans;
ll Y(int i){ return s[i]*i-dp[i];}
ll g(int i,int j){ return dp[i]-dp[j]-j*(s[i]-s[j]); }
int main()
{
    int N,i,j;
    scanf("%d",&N);
    for(i=1;i<=N;i++){
        scanf("%I64d",&s[i]);
        dp[i]=dp[i-1]+s[i]*i; 
        s[i]+=s[i-1]; 
        ans=max(ans,dp[i]);
    }
    for(i=1;i<=N;i++){
        int L=0,R=top-1,Mid,t=top;
        while(L<=R){
            Mid=(L+R)>>1;
            if(g(i,q[Mid])>=g(i,q[Mid+1])) t=Mid,R=Mid-1;
            else L=Mid+1;//或者用斜率二分 
        }
        ans=max(ans,g(i,q[t]));
        while(top&&(Y(i)-Y(q[top]))*(q[top]-q[top-1])>(Y(q[top])-Y(q[top-1]))*(i-q[top])) top--;
        q[++top]=i;
    }
    printf("%I64d\n",ans);
    return 0;
}