To solve this problem, we need to determine if a given list of integers is strictly increasing. A list is strictly increasing if each element is greater than the previous one.

Approach

1. Trivial Cases: A list with 0 or 1 elements is always strictly increasing.
2. Check Consecutive Pairs: For lists with more than one element, iterate through the list and check if each element is strictly less than the next. If any element is not strictly less than the next, the list is not strictly increasing.

Solution Code

def is_increasing(nums):
    if len(nums) <= 1:
        return True
    for i in range(len(nums)-1):
        if nums[i] >= nums[i+1]:
            return False
    return True

nums = list(map(int, input().split()))
result = is_increasing(nums)
print(result)

Explanation

1. Function is_increasing:

   • Trivial Case Handling: If the list length is ≤1, return True immediately.
   • Loop Through Pairs: The loop runs from 0 to len(nums)-2 (since we check each element with the next one). For each index i, if nums[i] is greater than or equal to nums[i+1], return False.
   • Return Result: If all pairs are strictly increasing, return True.

2. Main Execution:

   • Read input as a list of integers.
   • Call is_increasing on the list.
   • Print the result (either True or False).

This approach efficiently checks the strictly increasing property with a time complexity of O(n), where n is the length of the list, as it only iterates through the list once. The space complexity is O(1) since no additional space is used except for variables.

Answer: The filled blanks are len(nums)-1 and >=. The complete code is as above.

作者声明:本文包含人工智能生成内容。