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)
注：每块土地的三个参数：x , y , z。
x代表这块土地所占空间;
y代表这块土地上的动物数;
z代表这块土地的经济效益;
根据题意：每只动物所需支付的金钱为：x/y*z;
即所有动物所需金钱为：x/y*z*y=x*z;
代码：
#include "stdio.h" int main()
{
long long sum,x,y,z;
int n,f;
scanf("%d",&n);
while(n--)
{
sum=;
scanf("%d",&f);
while(f--)
{
scanf("%lld %lld %lld",&x,&y,&z);
sum=sum+x*z;
}
printf("%lld\n",sum);
}
return ;
}