1011. World Cup Betting (20)(最大值)
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
这题的IO例子纯坑爹! 例子里面输出百分位是四舍五入的,我一开始没注意,后来很高兴的注意到了,把它四舍五入了,然后就怎么都WA了。
最后试着把 +0.005 去掉,竟然AC了
AC代码:
#include <iostream>
#include <iomanip>
#include <string>
using namespace std;
double a[][];
int main()
{
while(cin>>a[][])
{
cin>>a[][]>>a[][];
int i,j;
for(i=;i<=;i++)
for(j=;j<=;j++)
cin>>a[i][j];
double max[]={,,};
string cc[];
for(i=;i<=;i++)
{
int temp;
for(j=;j<=;j++)
{
if(a[i][j]>max[i])
{
max[i]=a[i][j];
temp=j;
}
}
if(temp==)
{
cc[i]="W";
}
if(temp==) cc[i]="T";
if(temp==) cc[i]="L";
}
for(i=;i<=;i++)
cout<<cc[i]<<" ";
double sum=1.0;
for(i=;i<=;i++)
sum=sum*max[i];
sum=sum*0.65;
sum=(sum-)*;
cout<<fixed<<setprecision()<<sum<<endl;
}
return ;
}
1011. World Cup Betting (20)(最大值)的更多相关文章
- PAT 甲级 1011 World Cup Betting (20)(代码+思路)
1011 World Cup Betting (20)(20 分) With the 2010 FIFA World Cup running, football fans the world over ...
- PAT 甲级 1011 World Cup Betting (20)(20 分)
1011 World Cup Betting (20)(20 分)提问 With the 2010 FIFA World Cup running, football fans the world ov ...
- PAT甲 1011. World Cup Betting (20) 2016-09-09 23:06 18人阅读 评论(0) 收藏
1011. World Cup Betting (20) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Wit ...
- PAT 甲级 1011 World Cup Betting (20)(20 分)(水题,不用特别在乎精度)
1011 World Cup Betting (20)(20 分) With the 2010 FIFA World Cup running, football fans the world over ...
- PATA 1011 World Cup Betting (20)
1011. World Cup Betting (20) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Wit ...
- PAT (Advanced Level) Practice 1011 World Cup Betting (20 分) 凌宸1642
PAT (Advanced Level) Practice 1011 World Cup Betting (20 分) 凌宸1642 题目描述: With the 2010 FIFA World Cu ...
- 1011 World Cup Betting (20)(20 point(s))
problem With the 2010 FIFA World Cup running, football fans the world over were becoming increasingl ...
- PAT (Advanced Level) Practice 1011 World Cup Betting (20 分) (找最值)
With the 2010 FIFA World Cup running, football fans the world over were becoming increasingly excite ...
- PAT Advanced 1011 World Cup Betting (20 分)
With the 2010 FIFA World Cup running, football fans the world over were becoming increasingly excite ...
随机推荐
- 使用Hibernate框架技术时,对项目进行的配置
1.在需要使用Hibernate框架技术的项目上单击鼠标右键,在弹出的菜单中选择MyEclipse-->Add Hibernate Capabilities,打开Add Hibernate Ca ...
- http状态代码含义表
100 - 表示已收到请求的一部分,正在继续发送余下部分. 101 - 切换协议. 2xx - 成功.服务器成功地接受了客户端请求: 200 - 确定.客户端请求已成功. 201 - 已创建. 202 ...
- ORACLE之PACKAGE-游标变量
刚学pl/sql编程,写了两个package.pkg_temp_fn31和pkg_temp_fn32.内容涉及pl/sql基本语法,游标变量,存储过程(in,out). pkg_temp_fn31调用 ...
- Spring 数据源配置三:多数据源
在上一节中,我们讲述了多数据的情况: 1. 数据源不同(数据库厂商不同, 业务范围不同, 业务数据不同) 2. SQL mapper 文件不同, 3. mybatis + 数据方言不同 即最为简单的多 ...
- WP8.1 RSA 加解密实例(导入公钥私钥)
因项目上需要用到,之前在WP8.0的环境上调试通过,现在在开发8.1时发现已不支持原来的加密库,所以无法使用以前的方法,不得已,去寻找windows命名空间下RSA的加解密方法,经过几天的尝试,将解决 ...
- g++ 编译和链接(转)
传统意义上的编译程序分两步走 —— 编译和链接: 1.编译(compile):指用编译器(compiler)将源代码(source code)生成二进制目标文件(object file),在Windo ...
- 每天一道LeetCode--119.Pascal's Triangle II(杨辉三角)
Given an index k, return the kth row of the Pascal's triangle. For example, given k = 3,Return [1,3, ...
- hive 未初始化元数据库报错
启动hive-metastore和hive-server2 用beeline连接hive报错 [root@node04 hive]# beeline Beeline version 0.13.1-cd ...
- 申请Android Map 的API Key(v2)的最新申请方式(SHA1密钥)
申请Android Map 的API Key(v2)的最新申请方式(SHA1密钥)具体步骤如下: ...
- Learn Python The Hard Way学习笔记001
今天搜索了一下raw_input() 和 input()的区别,引用下原文部分内容 两个函数均能接收 字符串 ,但 raw_input() 直接读取控制台的输入(任何类型的输入它都可以接收).而对于 ...