Hey there! Let's tweak your Python program to find both the greatest and second-greatest number from those three inputs you're working with. I'll stick with the conditional statement style you're already using—no fancy built-in shortcuts, just straightforward checks that match your original code's vibe.

Here's the adjusted code that does exactly what you need:

x = int(input("Enter the 1st number: \n"))
y = int(input("Enter the 2nd number: \n"))
z = int(input("Enter the 3rd number: \n"))

# Step 1: Figure out which is the greatest number
if x > y and x > z:
    greatest = x
    # Now check the remaining two for second greatest
    second_greatest = y if y > z else z
elif y > x and y > z:
    greatest = y
    second_greatest = x if x > z else z
else:
    greatest = z
    second_greatest = x if x > y else y

# Output both results
print(f"{greatest} is the greatest number")
print(f"{second_greatest} is the second greatest number")

How this works: #

• First, we use chained conditionals to identify which of the three numbers is larger than the other two—this gives us our maximum value.
• Once we know the greatest number, the second largest has to be the bigger of the two leftover numbers. We just do a quick check between those two to get our runner-up.

If you prefer a slightly more compact version (still using only conditionals), you could write it like this instead:

x = int(input("Enter the 1st number: \n"))
y = int(input("Enter the 2nd number: \n"))
z = int(input("Enter the 3rd number: \n"))

# Calculate greatest with ternary conditionals
greatest = x if (x > y and x > z) else (y if y > z else z)

# Calculate second greatest based on which number was the max
second_greatest = y if (greatest == x and y > z) else (z if greatest == x else x if (greatest == y and x > z) else z if greatest == y else x if x > y else y)

print(f"{greatest} is the greatest")
print(f"{second_greatest} is the second greatest")

This does the exact same job, just wraps some of the checks into one-liner conditionals. Either version will work perfectly for your requirement.

内容的提问来源于stack exchange，提问作者Bhaskar baswala