[UVA10300]Ecological Premium

German farmers are given a premium depending on the conditions at their farmyard. Imagine the following simplied 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 nal premium of a farmer just multiply this premium per animal with the number of animals the farmer owns.

Input

The rst 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 farmers environment-friendliness. Input is terminated by end of le. 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 Ger- many’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

題意: 計算某個值，公式已給你了，仔細研究會發現只是第一項*第三項的總和，水題。

#include <stdio.h> 
#include <stdlib.h>  
int main() { 
  int n; 
  scanf("%d",&n); 
  while(n--) { 
    int m,size,num,cof,ans = 0; 
    scanf("%d",&m); 
    while(m--) { 
      scanf("%d %d %d",&size,&num,&cof); 
      ans += size * cof; 
    } 
    printf("%d\n",ans); 
  } 
  return 0; 
}