UVa OJ 10300

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

---

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

 #include <stdio.h>
 int main()
 {
 #ifdef CLOCK
     freopen("data.in", "r", stdin);
     freopen("data.out", "w", stdout);
 #endif
     int n,i,f,j,sum = ;
     int input[];
     scanf("%d", &n);
     for (i = ; i < n; i++)
     {
         if (n < )
             scanf("%d", &f);
         if (f >  && f < )
         for (j = ; j < f; j++)
         {
             scanf("%d%d%d", &input[], &input[], &input[]);
             sum += (input[] * input[]);
         }
         printf("%d\n", sum);
         sum = ;
     }
     return ;
 }

报告：还是一轮AC，不过用了重定向，以前没发现重定向非常方便去测验，傻傻的一直慢慢输，不过题目还真是一知半解，baidu后才知道大概，英语有待加强。

问题：无

解决：无