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 onthe conditions at their farmyard. Imagine the following simplified regulation:you know the size of each farmer's farmyard in square meters and the number ofanimals living at it. We won't make a difference between different animals,although this is far from reality. Moreover you have information about thedegree the farmer uses environment-friendly equipment and practices, expressedin a single integer greater than zero. The amount of money a farmer receivescan be calculated from these parameters as follows. First you need the space asingle animal occupies at an average. This value (in square meters) is thenmultiplied by the parameter that stands for the farmer'senvironment-friendliness, resulting in the premium a farmer is paid per animalhe owns. To compute the final premium of a farmer just multiply this premiumper animal with the number of animals the farmer owns.
Input
Thefirst line of input contains a single positive integer n (<20), the number of test cases. Each test case starts with aline containing a single integer f(0 
Output
Foreach test case output one line containing a single integer that holds thesummed burden for Germany's budget, which will always be a whole number. Do notoutput 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

SampleOutput
38
86
7445
(The Joint Effort Contest, Problem setter: Frank Hutter)
 
标题:1行目に1つのテスト用例の数(仮定m)を入力し、2行目に農夫のテスト用例の数(仮定n)を入力し、次にn行データの入力を行い、1行あたり3つのデータ、1つ目のデータは農夫が所有する土地面積、2つ目のデータは農夫が所有する動物の数、3つ目は生態係数である.計算式は土地面積/動物数です× せいたいけいすう×動物の数
MACコードは以下の通りである.

#include <stdio.h>

int main(void)
{
	int m, n;
	int a, b, c, sum;
	int i, j;

	while (scanf("%d", &m) != EOF) {
		for (i=0; i<m; ++i) {
			scanf("%d", &n);
			sum = 0;
			for (j=0; j<n; ++j) {
				scanf("%d%d%d", &a, &b, &c);
				sum += a * c;
			}	
			printf("%d
", sum);
		}
	}
	return 0;
}