Feel Good(两遍单调栈维护区间+前缀和）

最新推荐文章于 2022-01-14 11:44:48 发布

原创最新推荐文章于 2022-01-14 11:44:48 发布 · 375 阅读

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本文介绍了一种新颖的数学理论，该理论将人类情绪量化为非负整数，通过计算特定时间段内这些值的加权和来评估生活的总体质量。具体而言，时间段的价值由该段日子的情绪值之和与其最小情绪值的乘积决定。文章还提供了一个具体的算法实现案例，帮助读者理解如何寻找某人生命中最具价值的时间段。

Bill is developing a new mathematical theory for human emotions. His recent investigations are dedicated to studying how good or bad days influent people's memories about some period of life.

A new idea Bill has recently developed assigns a non-negative integer value to each day of human life.

Bill calls this value the emotional value of the day. The greater the emotional value is, the better the daywas. Bill suggests that the value of some period of human life is proportional to the sum of the emotional values of the days in the given period, multiplied by the smallest emotional value of the day in it. This schema reflects that good on average period can be greatly spoiled by one very bad day.

Now Bill is planning to investigate his own life and find the period of his life that had the greatest value. Help him to do so.

Input

The first line of the input contains n - the number of days of Bill's life he is planning to investigate(1 <= n <= 100 000). The rest of the file contains n integer numbers a1, a2, ... an ranging from 0 to 10 6 - the emotional values of the days. Numbers are separated by spaces and/or line breaks.

Output

Print the greatest value of some period of Bill's life in the first line. And on the second line print two numbers l and r such that the period from l-th to r-th day of Bill's life(inclusive) has the greatest possible value. If there are multiple periods with the greatest possible value,then print any one of them.

Sample Input

6
3 1 6 4 5 2

Sample Output

60
3 5

代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<map>
#include<set>
#include<stack>
#include<queue>
#include<vector>
#include<cmath>

const int maxn=1e5+5;
typedef long long ll;
using namespace std;
ll a[maxn];
ll sum[maxn];
ll L[maxn];
ll L2[maxn];
int main()
{
    int n;ll mn=0x3f3f3f3f;
    while(cin>>n)
    {
	for(int t=1;t<=n;t++)
	{
	    scanf("%lld",&a[t]);	
	    mn=min(mn,a[t]);
	}
	memset(sum,0,sizeof(sum));	
	for(int t=1;t<=n;t++)
	{
		sum[t]=sum[t-1]+a[t];
	}
	stack<int> S1,S2;
	while(!S1.empty())
	{
		S1.pop();
	}
	while(!S2.empty())
	{
		S2.pop();
	}

	for(ll t=1;t<=n;t++)
	{
        while(S1.size() && a[S1.top()] >= a[t]) S1.pop();
        if(S1.empty())     L[t] = 0;
        else              L[t] = S1.top();
        S1.push(t);
	}
	for(ll t=n;t>=1;t--)
	{
		while(S2.size() && a[S2.top()] >= a[t]) S2.pop();
        if(S2.empty())     L2[t] = n+1;
        else              L2[t] = S2.top();
        S2.push(t);
	}
	ll maxnn=n*mn;//注意这个初始化，很细节 
	int j=1,k=n;
	for(int t=1;t<=n;t++)
	{
		if((sum[L2[t]-1]-sum[L[t]])*a[t]>maxnn)
		{
			maxnn=(sum[L2[t]-1]-sum[L[t]])*a[t];
			j=L[t]+1;
			k=L2[t]-1;
		}
	}
	cout<<maxnn<<endl;
	cout<<j<<" "<<k<<endl;
    }
	return 0;
}