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最新推荐文章于 2018-02-12 05:56:41 发布

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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

1 1 1

2 2 2

3 3 3

2 3 4

8 9 2

9 1 8

6 12 1

8 1 1

10 30 40

9 8 5

100 1000 70

Sample Output

7445

代码：

#include <stdio.h>
#include <stdlib.h>
int a[40],c[40];
int main()
{
    int n;
    int m,b;
    long int count=0;
    int i,j,k;
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    {
        memset(a,0,sizeof(a));
        memset(c,0,sizeof(c));
        scanf("%d",&m);
        for(j=0;j<m;j++)
            scanf("%d%d%d",&a[j],&b,&c[j]);
        for(j=0;j<m;j++)
            count+=a[j]*c[j];
        printf("%ld\n",count);
        count=0;
    }
    return 0;
}