1011 World Cup Betting (20 point(s))
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 3.1 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 (4.1×3.1×2.5×65%−1)×2=39.31yuans (accurate up to 2 decimal places).
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
.
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.
1.1 2.5 1.7
1.2 3.1 1.6
4.1 1.2 1.1
T T W 39.31
Although this title did hard to understand. But it's easy to solve it. Certainly nothing to say. Look the code~
(The purpose of using English to portray my solution is that to exercise the ability of my expression of English and accommodate PAT advanced level's style.We can make progress together by reading and comprehending it. Please forgive my basic grammar's and word's error. Of course, I would appreciated it if you can point out my grammar's and word's error in comment section.( •̀∀•́ ) Furthermore, Big Lao please don't laugh at me because I just a English beginner settle for CET6 _(:з」∠)_ )
#include
int main()
{
int n;
char s[3]={'W','T','L'};
double m[3]={-1},t;
for(int i=0;i<3;++i)
{
int index=0;
double maxp=-1;
for(int j=0;j<3;++j)
{
scanf("%lf",&t);
if(t>maxp)
{
maxp=t;
index=j;
}
}
m[i]=maxp;
printf("%c ",s[index]);
}
double ans=(m[0]*m[1]*m[2]*0.65-1)*2;
printf("%.2f\n",ans);
return 0;
}