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

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

C++语言: Codee#23765

01 #include <iostream>
02 #include <cstdio>
03 using namespace std;
04
05 int main()
06 {
07
08     int a, b, c;
09     for(int n; scanf("%d", &n) != EOF;)
10         while(n--)
11         {
12             int m;
13             scanf("%d", &m);
14             int ans = 0;
15             while(m--)
16             {
17                 scanf("%d%d%d", &a, &b, &c);
18                 ans += a * c;
19             }
20             printf("%d\n", ans);
21         }
22
23
24     return 0;
25 }