POJ 题目2796 Feel Good（动态规划）

Feel Good

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 11041 Accepted: 3020Case 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

63 1 6 4 5 2

Sample Output

603 5

Source

Northeastern Europe 2005

题目大意：给你n个数的数组arr[n]。要你找到一个L和R是的。(arr[L]+.....+arr[R])*arr[k]的值最大。arr[k]为L->R的数的最小值。

分析：跟杭电oj1506差不多

ac代码

#include<stdio.h>#include<string.h>int l[100100],r[100100];__int64 a[100100],sum[100100];int main(){int n;scanf("%d",&n);int i;sum[0]=0;for(i=1;i<=n;i++){scanf("%I64d",&a[i]);sum[i]=sum[i-1]+a[i];l[i]=r[i]=i;}*for(i=1;i<=n;i++){while(l[i]>1&&a[l[i]-1]>=a[i]){l[i]=l[l[i]-1];}}for(i=n;i>=1;i--){while(r[i]<n&&a[r[i]+1]>=a[i]){r[i]=r[r[i]+1];}}__int64 ans=-1;int x,y;for(i=1;i<=n;i++){if((sum[r[i]]-sum[l[i]-1])*a[i]>ans){ans=(sum[r[i]]-sum[l[i]-1])*a[i];x=l[i];y=r[i];}}printf("%I64d\n%d %d\n",ans,x,y);}