PAT 1011
1011. World Cup Betting (20)
With the 2010 FIFA World Cup running, football fans the world over were becoming increasingly excited as the best players from the best teams doing battles for the World Cup trophy in South Africa. Similarly, football betting fans were putting their money where their mouths were, by laying all manner of World Cup bets.
Chinese Football Lottery provided a "Triple Winning" game. The rule of winning was simple: first select any three of the games. Then for each selected game, bet on one of the three possible results -- namely W for win, T for tie, and L for lose. There was an odd assigned to each result. The winner's odd would be the product of the three odds times 65%.
For example, 3 games' odds are given as the following:
W T L
1.1 2.5 1.7
1.2 3.0 1.6
4.1 1.2 1.1
To obtain the maximum profit, one must buy W for the 3rd game, T for the 2nd game, and T for the 1st game. If each bet takes 2 yuans, then the maximum profit would be (4.1*3.0*2.5*65%-1)*2 = 37.98 yuans (accurate up to 2 decimal places).
Input
Each input file contains one test case. Each case contains the betting information of 3 games. Each game occupies a line with three distinct odds corresponding to W, T and L.
Output
For each test case, print in one line the best bet of each game, and the maximum profit accurate up to 2 decimal places. The characters and the number must be separated by one space.
Sample Input
1.1 2.5 1.7
1.2 3.0 1.6
4.1 1.2 1.1
Sample Output
T T W 37.98
代码
1 #include <stdio.h>
2
3 int findMaxIndex(double*,int);
4 int main()
5 {
6 double odd[],profit;
7 int index1,index2,index3;
8 const char outPut[] = {'W','T','L'};
9 while(scanf("%lf%lf%lf",&odd[],&odd[],&odd[]) != EOF){
index1 = findMaxIndex(odd,);
profit = odd[index1];
scanf("%lf%lf%lf",&odd[],&odd[],&odd[]);
index2 = findMaxIndex(odd,);
profit *= odd[index2];
scanf("%lf%lf%lf",&odd[],&odd[],&odd[]);
index3 = findMaxIndex(odd,);
profit *= odd[index3];
printf("%c ",outPut[index1]);
printf("%c ",outPut[index2]);
printf("%c ",outPut[index3]);
profit = (profit * 0.65 - ) * ;
printf("%.2lf\n",profit);
}
return ;
}
int findMaxIndex(double *dArray,int n)
{
if(n == )
return -;
int index = ;
double maxValue = dArray[];
int i;
for(i=;i<n;++i){
if(maxValue < dArray[i]){
index = i;
maxValue = dArray[i];
}
}
return index;
}
PAT 1011的更多相关文章
- PAT 1011 World Cup Betting
1011 World Cup Betting (20 分) With the 2010 FIFA World Cup running, football fans the world over w ...
- PAT 1011 A+B和C (15)(C++&JAVA&Python)
1011 A+B和C (15)(15 分) 给定区间[-2^31^, 2^31^]内的3个整数A.B和C,请判断A+B是否大于C. 输入格式: 输入第1行给出正整数T(<=10),是测试用例的个 ...
- pat 1011 World Cup Betting(20 分)
1011 World Cup Betting(20 分) With the 2010 FIFA World Cup running, football fans the world over were ...
- PAT 1011. A+B和C (15)
给定区间[-231, 231]内的3个整数A.B和C,请判断A+B是否大于C. 输入格式: 输入第1行给出正整数T(<=10),是测试用例的个数.随后给出T组测试用例,每组占一行,顺序给出A.B ...
- 浙大pat 1011题解
With the 2010 FIFA World Cup running, football fans the world over were becoming increasingly excite ...
- PAT 1011 A+B和C
https://pintia.cn/problem-sets/994805260223102976/problems/994805312417021952 给定区间[-2^31^, 2^31^]内的3 ...
- PAT——1011. A+B和C
给定区间[-231, 231]内的3个整数A.B和C,请判断A+B是否大于C. 输入格式: 输入第1行给出正整数T(<=10),是测试用例的个数.随后给出T组测试用例,每组占一行,顺序给出A.B ...
- PAT 1011 World Cup Betting 查找元素
With the 2010 FIFA World Cup running, football fans the world over were becoming increasingly excite ...
- PAT 1011 World Cup Betting (20分) 比较大小难度级别
题目 With the 2010 FIFA World Cup running, football fans the world over were becoming increasingly exc ...
随机推荐
- 选择select框跳出信息
<html > <body > <select type="select" name=s1 onChange=alert("你选择了&quo ...
- Web API 依赖注入与扩展
与 MVC 类似, Web API 提供了System.Web.Http.Services.IDependencyResolver 接口来实现依赖注入, 我们可以很容易的用 Unity 来实现这个接口 ...
- MVC的EF编辑,不用查询直接修改
EF中会为每个 管理的 实体对象 创建一个代理包装类对象,其中会跟踪 实体对象 的状态和每个属性的状态: 一.通常使用EF更新的方式,先查询出要修改的数据,然后再修改新的值:实体对象被修改的属性 在 ...
- C# 中的 null
原文 C# 中的 null C# 允许将 null 值赋给任意引用变量(不能把 null 赋给一个值变量).值为 null 的变量表明该变量不引用内存中的任何对象. 如下所示: Circle c = ...
- MemoryMappedFile 内存映射文件 msdn
http://msdn.microsoft.com/zh-cn/library/dd997372%28v=vs.110%29.aspx 内存映射文件 .NET Framework 4.5 其他版本 1 ...
- alibaba笔试1
5.D 一个线程不可以改变另一个线程的程序计数器.如果改变了,线程在切换后就恢复不到正确的位置. 一个线程可以访问另一个线程的栈.http://bbs.csdn.net/topics/39008942 ...
- Flash 导出图片和声音
命令文件 PolarBear_jsfl.zip Flash Professional 编辑器命令,用来导出 flash 库中的图片和声音 使用步骤: 1. 首先下载 PolarBear_jsfl.zi ...
- Ubuntu 14.04 设置静态IP
使用Network Manager UI界面中指定 手动时,无法保存. 通过修改配置文件解决来此问题.记录以下. 如果输入过密码后,就会出现在这个目录下面, 以如下chinaNet为例 gaojing ...
- 如何实现CSS居中?–CSS居中常用方法
来源:http://www.ido321.com/824.html 一.水平居中 1.内联元素居中:相对父级块级元素居中对齐 1: .center-children { 2: text-align: ...
- Android 网络权限配置
Android开发应用程序时,如果应用程序需要访问网络权限,需要在 AndroidManifest.xml 中加入以下代码 <uses-permission android:name=”andr ...