POJ2796 Feel Good【单调栈+前缀和 非负区间最小值乘区间和】

Feel Good

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 20284		Accepted: 5597
Case Time Limit: 1000MS		Special Judge

Description

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 106 - 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

Source

Northeastern Europe 2005

题意

给定一个非负数组，求区间最小值乘区间和最大值，输出这个最大值，并记录相应的左右闭区间的端点值，如果有多个区间，则输出任意一个

思路

由于是非负数组，因此区间越大，其区间和才可能越大，枚举每个数，使用单调栈计算以它为最小值向左向右扩展的最大区间，不断更新答案即可。

C++代码

#include

using namespace std;

const int N=100005;

typedef long long ll;

ll sum[N];
int a[N],L[N],s[N],l,r;

int main()
{
	int n;
	while(~scanf("%d",&n))
	{
		sum[0]=0;
		for(int i=1;i<=n;i++)
		{
			scanf("%d",&a[i]);
			L[i]=i;
			sum[i]=sum[i-1]+a[i];
		}
		a[n+1]=-1;//添加一个最小值，保证能把栈清空 
		int top=0;
		ll ans=-1;
		//维护一个单点递增栈 
		for(int i=1;i<=n+1;i++)
		{
			while(top>0&&a[s[top]]>a[i])
			{
				ll temp=(sum[i-1]-sum[L[s[top]]-1])*a[s[top]];
				if(ans0&&a[s[top]]==a[i]) L[i]=L[s[top]];
			s[++top]=i;
		}
		printf("%lld\n%d %d\n",ans,l,r);
	}
	return 0;
}