UVA 10300 - Ecological Premium

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 integerf (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 than100000 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

 

 

题目很长，题意也兜来兜去。但很容易发现动物数量n其实是没用的，先除了后乘回，这也可以在题目中体现，因为他说保证数字都是整数。结果每人就是 s*d

code：

#include<stdio.h>
int main()
{
    int i,t,f,s,n,d,result;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&f);
        result=0;
        for(i=0;i<f;i++)
        {
            scanf("%d%d%d",&s,&n,&d);
            result+=s*d;
        }
        printf("%d\n",result);
    }
    return 0;
}