Solution of 10300 - Ecological Premium

Problem Description
source:https://uva.onlinejudge.org/external/103/10300.html

German farmers are given a premium depending on the conditions at their farmyard. Imagine the following simplified regulation: you know the size of each farmer’s farmyard in square meters and the number of animals living at it. We won’t make a difference between different animals, although this is far from reality. Moreover you have information about the degree the farmer uses environment-friendly equipment and practices, expressed in a single integer greater than zero. The amount of money a farmer receives can be calculated from these parameters as follows. First you need the space a single animal occupies at an average. This value (in square meters) is then multiplied by the parameter that stands for the farmer’s environment-friendliness, resulting in the premium a farmer is paid per animal he owns. To compute the final premium of a farmer just multiply this premium per animal with the number of animals the farmer owns.

Input 

The first line of input contains a single positive integer n (< 20), the number of test cases. Each test case starts with a line containing a single integer f (0 < f < 20), the number of farmers in the test case. This line is followed by one line per farmer containing three positive integers each: the size of the farmyard in square meters, the number of animals he owns and the integer value that expresses the farmers environment-friendliness. Input is terminated by end of file. No integer in the input is greater than 100000 or less than 0.

Output 

For each test case output one line containing a single integer that holds the summed burden for Germany’s budget, which will always be a whole number. Do not output any blank lines.  

Sample Input 

3 
5 
1 1 1 
2 2 2 
3 3 3 
2 3 4 
8 9 2 
3 
9 1 8 
6 12 1 
8 1 1 
3 
10 30 40
 9 8 5 
100 1000 70 

Sample Output 

38 
86 
7445

Solution:

#include<stdio.h>

int main()
{
    long int testcase, n, a, b, c,sum,i,j;
    while(scanf("%ld",&testcase)==1)
    {
        for(i=1;i<=testcase;i++)
        {
            scanf("%ld",&n);
            sum=0;
            for(j=1;j<=n;j++)
            {
                scanf("%ld%ld%ld",&a,&b,&c);
                sum+=(a*c);
            }
            printf("%ld\n",sum);
        }
    }
    return 0;

}