CareerCup

x>y ? x>z ? x : z : y > z ? y : z

(x>y?x:y)>z?(x>y?x:y):z

(((x>y)?x:y)>z)?((x>y)?x:y):z

(x>y)? ( (x > z) ? x : z ) : ( (y > z) ? y : z )

max = num1 > num2 ? num1 > num3? num1 : num3: num2 > num3 ? num2:num3;

we can use concept
suppose if b>a then below concept can give b as output
c is average of a and b(in between) ;
d---> distance between avg and number(a or b);
c=(a+b)/2;
d=abs(a-b)/2;
so max number will be (c+d)..

a--------c--------b
<---d--->|<--d---->
so max that is b is (c+d);
for three number now get between b and another number using same concept.

I think the point of the question is to use conditional operator only ONCE.
I can do it using it twice:
( x > y && y > z ) ? x : ( y > z ? y : z );

int a,b,c,d;
  cin>>a>>b>>c;
  d = (a>b)?(a>c?a:c):(b>c?b:c);
  cout<<d<<endl;

all the above fail for cases where all numbers are equal. It would only output the one that was compared last in the conditional.

max=a>b?(c>a?c:a):(c>b?c:b);

int i=6,j=4,k=3,Max;
Max= i*(i>j&& i>k)+j*(j>i &&j> k)+k *(k>i&&k>j);

Do i miss , any special case??

int a=90,b=40,c=20;
int max;
max=((a >b)?((a>c)?a:c):((b>c)?b:c));