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Maximize the Earning

 Problem:-

Napoleon choosed a city for Advertising his company's product. There are S streets in that city. Each day he travels one street. There are N buildings in a street which are located at points 1,2,3....N(respectively). Each building has some height(Say h meters). Napoleon stands at point 0. His height is 0.5m. Now Napoleon starts communicating with the people of each building. He can communicate with people of a particular building only if he can see that building. If he succeed to communicate with any particular building then his boss gives him Rrupees. i.e. if he communicates with K buildings in a day, then he will earn KRrupees. Now Napoleon wants to know his maximum Earning for each day.

Note: All the points are on Strainght Line and Napoleon is always standing at point 0.

Input:
First line of input contains an integer S, denoting no of streets in the city.
Details for each street is described in next two lines.
First line contains two integers N and R, denoting no of buildings in the Street and earning on communicating with a building.
Second line contains N integers, denoting height of  building.

Output:
Print S Lines, each containing maximum earning in ith street.
    
Constraints:
    1N104
    1h109
    1S100
    1R104

Sample Input
2
6 3
8 2 3 11 11 10
5 12
12 20 39 45 89
Sample Output
6
60
Time Limit: 1
Memory Limit: 256
Source Limit:
Explanation

There are two streets in the city.

The first street has 6 buildings and the earning of Napoleon for communicating with each building is 3rupees.

Height of buildings are 8,2,3,11,11,10 respectively.

As Chef is standing at point 0, he will be able to see only 1st and 4th building.

So his total earning will be 32=6rupees in that street

Similarly for 2nd street his earning will be 60rupees


Code:-

#include<stdio.h>
int main()
{
    long int n,s,r,i;
    long long int max,h,building;
    scanf("%ld",&s);
    
    while(s--)
    {
        scanf("%ld%ld",&n,&r);
         max=0;
        building=0;
    for(i=0;i<n;i++)
        {
      scanf("%lld",&h);
      if(h>max)
           {
               max=h;
               building++;
           }
        }
        printf("%lld\n",r*building);

    }
    return 0;
}