A histogram is a polygon composed of a sequence of rectangles aligned at a common base line. The rectangles have equal widths but may have different heights. For example, the figure on the left shows the histogram that consists of rectangles with the heights 2, 1, 4, 5, 1, 3, 3, measured in units where 1 is the width of the rectangles:

Usually,
histograms are used to represent discrete distributions, e.g., the
frequencies of characters in texts. Note that the order of the
rectangles, i.e., their heights, is important. Calculate the area of the
largest rectangle in a histogram that is aligned at the common base
line, too. The figure on the right shows the largest aligned rectangle
for the depicted histogram.

 

The input contains several test cases. Each test case describes a
histogram and starts with an integer n, denoting the number of
rectangles it is composed of. You may assume that 1 <= n <=
100000. Then follow n integers h1, ..., hn, where 0 <= hi <=
1000000000. These numbers denote the heights of the rectangles of the
histogram in left-to-right order. The width of each rectangle is 1. A
zero follows the input for the last test case.

 

For each test case output on a single line the area of the largest
rectangle in the specified histogram. Remember that this rectangle must
be aligned at the common base line.

 

#include<stdio.h>

#include<string.h>

#include<iostream>

#include<algorithm>

using namespace std;

const int maxn=;

typedef long long int ll;

ll l[maxn],r[maxn],h[maxn];

int main(){

    int n;

    while(scanf("%d",&n)!=EOF){

        if(!n)

        break;

        memset(l,,sizeof(l));

        memset(r,,sizeof(r));

        memset(h,,sizeof(h));

  //          memset(dp,0,sizeof(dp));

        for(int i=;i<=n;i++){

           scanf("%lld",&h[i]);

        }

        l[]=;

        r[n]=n;

        for(int i=;i<=n;i++){

             int tmp=i;

            while(tmp>&&h[i]<=h[tmp-])

            tmp=l[tmp-];

            l[i]=tmp;

        }

        for(int i=n-;i>=;i--){

              int tmp=i;

            while(tmp<n&&h[i]<=h[tmp+])

            tmp=r[tmp+];

            r[i]=tmp;

        }

        ll ans=-;

        for(int i=;i<=n;i++){

           ll tmp=(r[i]-l[i]+)*h[i];

            ans=max(ans,tmp);

        }

      printf("%lld\n",ans);

    }

    return ;

}