1011 World Cup Betting

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=39.31 yuans (accurate up to 2 decimal places).

Sample Input:

1.1 2.5 1.7
1.2 3.1 1.6
4.1 1.2 1.1

Sample Output:

T T W 39.31

#include<cstdio>
#include<string>
#include<algorithm>
using namespace std;

struct odds
{
  float w;
  float t;
  float l;
  float max;
}games[3];

float Max(odds item)
{
  if (item.w > item.t && item.w > item.l)
    return item.w;
  else if (item.t > item.l)
    return item.t;
  else
    return item.l;
}

int main()
{
  float re = 0.65;
  char re_c[3];
  for (int i = 0; i < 3; i++)
  {
    scanf_s("%f%f%f", &games[i].w, &games[i].t, &games[i].l);
    if (games[i].w > games[i].t && games[i].w > games[i].l)
    {
      re_c[i] = 'W';
      re *= games[i].w;
    }
    else if (games[i].t > games[i].l)
    {
      re_c[i] = 'T';
      re *= games[i].t;
    }
    else
    {
      re_c[i] = 'L';
      re *= games[i].l;
    }
  }
  re = (re - 1) * 2;
  for (int i = 0; i < 3; i++)
  {
    printf("%c ", re_c[i]);
  }
  printf("%.2f\n", re);
}