Time Limit: 3000MS
Memory Limit: 65536K
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 $(1$

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

題意：
求一個區間，使得（區間和*區間最小值）最大
輸出ans和任意一個答案區間

題解：
列舉每個點，詢問將其當作區間最小值的最大區間和是多少
由於所有數字都是非負數，所以區間越長越好，記L[i]為第i個元素為最小值的區間最左端，記R[i]為第i個元素為最小值的區間最右端。用單調棧來解決這個問題。（正反各掃一遍）

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<stack>
#define LiangJiaJun main
#define ll long long
#define pa pair<int,int>
using namespace std;
int n,a[100004];
stack<pa>st;
ll s[100004];
int L[100004],R[100004];
int w33ha(){
    s[0]=0;
    while(!st.empty())st.pop();
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
        s[i]=s[i-1]+a[i];
    }
    for(int i=1;i<=n;i++){
        while(!st.empty()&&st.top().first>=a[i])st.pop();
        if(st.empty())L[i]=1;
        else L[i]=st.top().second+1;
        st.push(make_pair(a[i],i));
    }
    while(!st.empty())st.pop();
    for(int i=n;i>=1;i--){
        while(!st.empty()&&st.top().first>=a[i])st.pop();
        if(st.empty())R[i]=n;
        else R[i]=st.top().second-1;
        st.push(make_pair(a[i],i));
    }
    ll ans=-1;
    int ansl=0,ansr=0;
    for(int i=1;i<=n;i++){
        if(ans<(s[R[i]]-s[L[i]-1])*1LL*a[i]){
            ans=(s[R[i]]-s[L[i]-1])*1LL*a[i];
            ansl=L[i];
            ansr=R[i];
        }
    }
    printf("%lld\n%d %d\n",ans,ansl,ansr);
    return 0;
}
int LiangJiaJun(){
    while(scanf("%d",&n)!=EOF)w33ha();
    return 0;
}