2013秋13级预备队集训练习1 C - 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

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

#include<stdio.h>
int main()
{
	int i , n , j , f , a[20][3] , sum ;
	scanf("%d", &n);
	for(i = 1 ; i <= n ; i++)
	{
		scanf("%d", &f);
		sum = 0 ;
		for(j = 0 ; j < f ; j++)
		{
			scanf("%d %d %d", &a[j][0],&a[j][1],&a[j][2]);
				sum = sum + a[j][0]*a[j][2];
		}
		printf("%d\n", sum);
	}
}

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