10300 - Ecological Premium
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)
这道题说的是一个运算：总共有T组数据，每组数据中有多组case，每组case中第一个数除以第二个数再乘以第三个数，再乘第二个数。所以就是第一个数乘第三个数。看着说了好多，其实再考你的英文水平哈！
#include "stdio.h"
int main ()
{
int T;scanf("%d",&T);
while(T--)
{
int N,sum = 0;scanf("%d",&N);
for (int i=1;i<=N;i++)
{
int a,b,c;scanf("%d%d%d",&a,&b,&c);
sum+=a*c;
}
printf("%d\n",sum);
}
}
