uva 10300

Problem A

Ecological Premium

Input: standard input

Output: standard output

Time Limit: 1 second

Memory Limit: 32 MB

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 farmer’s 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

#include <stdio.h>
#include <string.h>
#define N 21
int a
;
int b
;
int c
;
int main()
{
int n, m, i;
int sum;
scanf("%d", &n);
while(n--)
{
sum = 0;
memset(a, 0, sizeof(a));
memset(b, 0, sizeof(b));
memset(c, 0, sizeof(c));
scanf("%d", &m);
for(i = 0; i < m; i++)
{
scanf("%d%d%d", &a[i], &b[i], &c[i]);
sum += a[i] * c[i];
}
printf("%d\n", sum);
}
return 0;
}