欢迎访问我的新博客:http://www.milkcu.com/blog/

原文地址:http://www.milkcu.com/blog/archives/uva10300.html

题目描述

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)

解题思路

已知农场的面积和动物的数目。

此外还知道农场主使用设备的环境友好等级。

农场主得到的奖金可以由上面三个参数计算得出。

首先获得每只动物平均占用的面积。

然后这个值乘以环境友好度参数,得到的是一只动物得到的奖金。

最后把上面的值乘以动物的数目,就可以得到农场主获得的奖金。



奖金 = 农场的面积 / 动物的数目 * 环境友好等级 * 动物的数目 = 农场的面积 * 环境友好等级 



No integer in the input is greater than 100000 or less than 0.

注意数值范围,需要用long long保存。

代码实现

#include <iostream>
using namespace std;
int main(void) {
int n;
cin >> n;
while(n--) {
int f;
cin >> f;
long long s = 0;
while(f--) {
long long a, b, c;
cin >> a >> b >> c;
s += a * c;
}
cout << s << endl;
}
return 0;
}

(全文完)

Ecological Premium - UVa10300的更多相关文章

  1. 10300 - Ecological Premium

    Problem A Ecological Premium Input: standard input Output: standard output Time Limit: 1 second Memo ...

  2. UVa 10300 - Ecological Premium

    https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=94&page=s ...

  3. Ecological Premium

    #include<bits/stdc++.h> using namespace std; int main() { int n,m; unsigned long long int a,b, ...

  4. UVA_10300:Ecological Premium

    Sample Input 351 1 12 2 23 3 32 3 48 9 239 1 86 12 18 1 1310 30 409 8 5100 1000 70Sample Output 3886 ...

  5. UVA题目分类

    题目 Volume 0. Getting Started 开始10055 - Hashmat the Brave Warrior 10071 - Back to High School Physics ...

  6. UVa OJ 10300

    Problem A Ecological Premium Input: standard input Output: standard output Time Limit: 1 second Memo ...

  7. 【索引】Volume 0. Getting Started

    AOAPC I: Beginning Algorithm Contests (Rujia Liu) Volume 0. Getting Started 10055 - Hashmat the Brav ...

  8. 数据库管理工具GUI - PremiumSoft Navicat Premium Enterprise 11.2.15 x86/x64 KEY

    转载自: 数据库管理工具GUI - PremiumSoft Navicat Premium Enterprise 11.2.15 x86/x64 KEY Navicat Premium(数据库管理工具 ...

  9. mysql 5.7 的安装配置与 navicat premium for mysql 11 的破解使用

    再安装mysql5.7 或以上的版本出现了一些问题,现在总结下,希望能给初入学习mysql的人一下帮助,大牛就不要来嘲笑小弟我了 首先准备如下: 1.下载mysql 5.7,下载地址:https:// ...

随机推荐

  1. Skynumber

    Time Limit: 1000ms Memory Limit: 128000KB 64-bit integer IO format:      Java class name: Submit Sta ...

  2. android开源系列:CircleImageView采用圆形控制它们的定义

    1.定义自己的圆形控制github住址:https://github.com/hdodenhof/CircleImageView 基本的类: package de.hdodenhof.circleim ...

  3. Nyoj 网络的可靠性(水题)

    描述 A公司是全球依靠的互联网解决方案提供商,也是2010年世博会的高级赞助商.它将提供先进的网络协作技术,展示其”智能+互联“的生活概念,同时为参观者提供高品质的个人体验和互动,以”信息通信,尽情城 ...

  4. C#精华(文章3版本)笔记

    C#精华(文章3版本) 跳转至: 导航. 搜索 文件夹 1 C#概述 2 数据类型 3 运算符和控制流 4 方法和參数 5 类 6 继承 7 接口 8 值类型(struct) 9 合式类型 10 异常 ...

  5. js之with

    网上找资料的时候看到有js代码里出现了大量的with语句,有点好奇,这里做下总结: 1)简要说明         with 语句可以方便地用来引用某个特定对象中已有的属性,但是不能用来给对象添加属性. ...

  6. Session与Caching

    Session与Caching 在之前的版本中,Session存在于System.Web中,新版ASP.NET 5中由于不在依赖于System.Web.dll库了,所以相应的,Session也就成了A ...

  7. PLSQL_监控有些SQL的执行次数和频率

    原文:PLSQL_监控有些SQL的执行次数和频率 2014-12-25 Created By 鲍新建

  8. 从WebBrowser中取得Cookie 和 WebClient设置cookie!

    原文:从WebBrowser中取得Cookie 和 WebClient设置cookie! 从WebBrowser中取得Cookie 的代码 CookieContainer myCookieContai ...

  9. ActivityLifeCycle官方demo分解

    1.左右Activity生命周期的若干条款: p=330">http://1.duoinfo.sinaapp.com/? p=330 http://1.duoinfo.sinaapp. ...

  10. MVC验证05-自定义验证规则、验证2个属性值不等

    原文:MVC验证05-自定义验证规则.验证2个属性值不等 本文体验2个属性值不等.即当一个属性输入值,另外一个属性输入的值不能和第一个属性值相等.相关文章包括: MVC验证01-基础.远程验证   M ...