| 题目描述 |
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
| 解题思路 |
一道水题，需要注意的是精度问题，对于保留两位小数无需多言，用printf保留两位输出即可，但是double类型在牛客网上会卡死在一个测试用例上，如果改成float类型即可通过．
| 代码设计 |
//部分代码
//zhicheng
char m[]={'W','T','L'};
int main()
{
float pro=1;
for(int i=0;i<3;i++)
{
float tmp,maxn=0;int maxn_id;
for(int j=0;j<3;j++){scanf("%f",&tmp);if(tmp>maxn) {maxn=tmp;maxn_id=j;}}
printf("%c ",m[maxn_id]);
pro*=maxn;
}
printf("%.2f\n",(pro*0.65-1)*2);
return 0;
}
有关PAT (Basic Level) 的更多内容可以关注 ——> PAT-B题解
有关PAT (Advanced Level) 的更多内容可以关注 ——> PAT-A题解