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)
Solution:
#include <stdio.h>
int main(){
static int T, N, S, A, F, P;
scanf("%d", &T);
while(T--){
scanf("%d", &N);
P = 0;
while(N--){
scanf("%d%d%d", &S, &A, &F);
P += S*F;
}
printf("%d\n", P);
}
return 0;
}
int main(){
static int T, N, S, A, F, P;
scanf("%d", &T);
while(T--){
scanf("%d", &N);
P = 0;
while(N--){
scanf("%d%d%d", &S, &A, &F);
P += S*F;
}
printf("%d\n", P);
}
return 0;
}
No comments:
Post a Comment