Input FormatThe first line contains a single integer, .ConstraintsOutput FormatIf , then print the lowercase English word corresponding to the number (e.g., one for , two for , etc.); otherwise, print Greater than 9 instead.Sample Input5Sample OutputfiveSample Input #018Sample Output #01eightSample Input #0244Sample Output #02Greater than 9
Question
Input FormatThe first line contains a single integer, .ConstraintsOutput FormatIf , then print the lowercase English word corresponding to the number (e.g., one for , two for , etc.); otherwise, print Greater than 9 instead.Sample Input5Sample OutputfiveSample Input #018Sample Output #01eightSample Input #0244Sample Output #02Greater than 9
Solution
This problem is asking for a program that takes an integer as input and outputs the English word for that number if it is less than or equal to 9. If the number is greater than 9, the program should output "Greater than 9".
Here is a step-by-step solution in Python:
- First, we need to create a list of English words for numbers from 1 to 9:
numbers = ["one", "two", "three", "four", "five", "six", "seven", "eight", "nine"]
- Then, we take an integer as input from the user:
n = int(input().strip())
- We check if the number is less than or equal to 9. If it is, we print the corresponding English word from our list. Remember that list indices in Python start at 0, so we need to subtract 1 from our number to get the correct index. If the number is greater than 9, we print "Greater than 9":
if n <= 9:
print(numbers[n-1])
else:
print("Greater than 9")
So, the complete code would be:
numbers = ["one", "two", "three", "four", "five", "six", "seven", "eight", "nine"]
n = int(input().strip())
if n <= 9:
print(numbers[n-1])
else:
print("Greater than 9")
Similar Questions
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