UVa 10300 Ecological Premium

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

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(The Joint Effort Contest, Problem setter: Frank Hutter)

#include <stdio.h>
int main(int argc, const char *argv[])
{
    int n, f, i, j, a, b, c, sum;
    scanf("%d", &n);
    for (i = 0; i < n; i++) {
        scanf("%d", &f);
        sum = 0;
        for (j = 0; j < f; j++) {
            scanf("%d%d%d", &a, &b, &c); 
            sum += a * c;
        }
        printf("%d\n", sum);
    }
    return 0;
}