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Time Limit:?3000MS	?	Memory Limit:?65536K
Total Submissions:?14435	?	Accepted:?3996
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 【解析】 題目大意：找一個區間，區間的和乘以區間的最小值最大。 暴力一定超時。 我們可以對于每一個元素a[i]，找到一個盡可能大的區間，使a[i]成為這個區間的最小值。 用sum數組存前綴和，最后枚舉區間即可。 注意long long 【code】 #include<iostream> #include<cstdio> using namespace std; #define N 100050 long long sum[N],l[N],r[N],a[N]; long long n,L,R; long long ans,maxx=-1; int main() {scanf("%d",&n);for(int i=1;i<=n;i++){scanf("%lld",&a[i]);l[i]=r[i]=i;sum[i]=sum[i-1]+a[i];}a[0]=a[n+1]=-1;for(int i=1;i<=n;i++){while(a[i]<=a[l[i]-1])l[i]=l[l[i]-1];}for(int i=n;i>=1;i--){while(a[i]<=a[r[i]+1])r[i]=r[r[i]+1];}for(int i=1;i<=n;i++){ans=(sum[r[i]]-sum[l[i]-1])*a[i];if(ans>maxx){maxx=ans;L=l[i];R=r[i];}}printf("%lld\n%lld %lld",maxx,L,R);return 0; }

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轉載于:https://www.cnblogs.com/zzyh/p/6852056.html

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