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

 

 

代码：

#include<stdio.h>

int main()

 {

    int
cases,num,a,b,c,i,sum;

 scanf("%d",&cases);

   
while(cases--)

    {

   
scanf("%d",&num);

    sum=0;

   
for(i=0;i<num;i++)

    {

    
scanf("%d %d
%d",&a,&b,&c);

    
sum+=a*c;

    }

   
printf("%d\n",sum);

 }

  return 0;

}