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)
1 #include <stdio.h> 2 int main() 3 { 4 #ifdef CLOCK 5 freopen("data.in", "r", stdin); 6 freopen("data.out", "w", stdout); 7 #endif 8 int n,i,f,j,sum = 0; 9 int input[5]; 10 scanf("%d", &n); 11 for (i = 0; i < n; i++) 12 { 13 if (n < 20) 14 scanf("%d", &f); 15 if (f > 0 && f < 20) 16 for (j = 0; j < f; j++) 17 { 18 scanf("%d%d%d", &input[0], &input[1], &input[2]); 19 sum += (input[0] * input[2]); 20 } 21 printf("%d\n", sum); 22 sum = 0; 23 } 24 return 0; 25 }
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
