UVA - 10300 Ecological Premium(水题)
2014-08-22 09:05
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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 integerf (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 farmers 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
Sample Output
38
86
7445
题意:
第一行输入一个测试用例的数量m,第二行输入农夫测试用例的数量n,
接着是n行数据的输入,每行3个数据,
分别是农夫拥有的土地面积、农夫拥有的动物的数量、生态系数。
要求你计算出农夫应获得的奖金。
计算公式:
奖金 = ∑((土地面积/动物数量)×生态系数×动物数量) = ∑(土地面积×生态系统数)
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 integerf (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 farmers 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
题意:
第一行输入一个测试用例的数量m,第二行输入农夫测试用例的数量n,
接着是n行数据的输入,每行3个数据,
分别是农夫拥有的土地面积、农夫拥有的动物的数量、生态系数。
要求你计算出农夫应获得的奖金。
计算公式:
奖金 = ∑((土地面积/动物数量)×生态系数×动物数量) = ∑(土地面积×生态系统数)
#include <stdio.h> int main() { int m,n; int a,b,c,sum; while(scanf("%d",&m) != EOF) { while(m--) { scanf("%d",&n); sum = 0; for(int i = 0; i < n; i++) { scanf("%d%d%d",&a,&b,&c); sum += a*c; } printf("%d\n",sum); } } return 0; }
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