2013秋13级预备队集训练习1 C - Ecological Premium
2013-12-09 08:51
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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
333
2 34
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() { int i , n , j , f , a[20][3] , sum ; scanf("%d", &n); for(i = 1 ; i <= n ; i++) { scanf("%d", &f); sum = 0 ; for(j = 0 ; j < f ; j++) { scanf("%d %d %d", &a[j][0],&a[j][1],&a[j][2]); sum = sum + a[j][0]*a[j][2]; } printf("%d\n", sum); } }
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