[PAT (Advanced Level) ]1011. World Cup Betting 解题文档
2015-10-17 14:09
423 查看
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
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
分析:很简单。直接上代码。如下。
#include<iostream>
#include<iomanip>
using namespace std;
int main(){
double matrix[3][3];
double result=0.65;
int flag[3]={0};
for(int i=0;i<3;i++){
double big;
cin>>matrix[i][0];
big=matrix[i][0];
for(int j=1;j<3;j++){
cin>>matrix[i][j];
if(big<matrix[i][j]){
big=matrix[i][j];
flag[i]=j;
}
}
result*=big;
}
for(int i=0;i<3;i++){
switch(flag[i])
{
case 0:{cout<<"W ";break;}
case 1:{cout<<"T ";break;}
case 2:{cout<<"L ";break;}
}
}
result=(result-1)*2;
cout<<fixed<<setprecision(2)<<result;
return 0;
}
相关文章推荐
- c总结2 ---自己实现字符串的拷贝(指针学习1)
- AsyncTask的用法
- 邮件服务器DNS设置-----MX、SPF、DKIM记录详解
- 小感悟
- Django1.8 关于 静态文件配置
- OBJ文件格式
- Js中获取超链接里面传递的参数值
- java
- poj(3254)——Corn Fields
- 闪回delete 恢复
- 也谈virtual
- leedcode: Course Schedule
- uint8_t 头文件为#include<stdint.h>
- 数组随机排序
- [Leetcode] Binary Tree Paths, Solution
- Merge Intervals
- 【图像处理】数字图像处理软件-特效--光晕特效(五)
- 黑马程序员--指向指针的指针 写的非常好 适合0基础菜鸟
- printf PRIu64
- 微信jsSDK开发