This problem requires writing an algorithm to sort N given numbers in ascending order. The range of the given numbers is from 1 to 10000, and each number is an integer greater than 0 and less than or equal to 10000. Therefore, we need to implement an efficient sorting algorithm considering this aspect.<\/p>\n
Please write a program that sorts the given N numbers and prints one number per line. The maximum number of inputs can be 1,000,000. Thus, since we need to sort a large quantity of numbers, we must consider the time complexity.<\/p>\n
The first line of the input contains the number of elements N (1 \u2264 N \u2264 1,000,000), followed by N lines each containing a number.<\/p>\n
Print the sorted numbers one per line.<\/p>\n
\n 5\n 5\n 4\n 3\n 2\n 1\n <\/pre>\nExample Output<\/h2>\n
\n 1\n 2\n 3\n 4\n 5\n <\/pre>\nProblem Analysis<\/h2>\n
Given the constraints, we can use a sorting algorithm that is efficient in terms of time and space. Python’s built-in sorting functionality utilizes Timsort, which has a time complexity of O(N log N) in the average and worst-case scenarios. However, considering the range of numbers is limited (from 1 to 10000), we can also explore counting sort or bucket sort for this specific task.<\/p>\n
Proposed Solution<\/h2>\n
We will use Counting Sort to solve this problem. Counting Sort works as follows:<\/p>\n
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- Create a counting array to count the number of input numbers. The size of the array is set to 10001 (including 0~10000).<\/li>\n
- Store the occurrence count of each number in the counting array.<\/li>\n
- Iterate through the counting array to print each number in sorted order.<\/li>\n<\/ul>\n
Code Implementation<\/h2>\n
\nimport sys\n\ndef counting_sort(n, numbers):\n count = [0] * 10001 # Create size 10001 for numbers range from 0 to 10000\n\n for num in numbers:\n count[num] += 1 # Count occurrences of the numbers\n\n result = []\n for i in range(10001):\n while count[i] > 0:\n result.append(i) # Add to result array based on count array values\n count[i] -= 1\n\n return result\n\nif __name__ == \"__main__\":\n n = int(sys.stdin.readline())\n numbers = [int(sys.stdin.readline()) for _ in range(n)]\n \n sorted_numbers = counting_sort(n, numbers)\n print(\"\\n\".join(map(str, sorted_numbers)))\n <\/pre>\nMain Points<\/h3>\n
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- Counting Sort is a stable sorting algorithm. This means that the relative position of elements with the same value is preserved.<\/li>\n
- This algorithm has a time complexity of O(N + K), where N is the size of the input and K is the range of the numbers.<\/li>\n
- The memory usage is also efficient as long as K is not too large.<\/li>\n<\/ul>\n
Conclusion<\/h2>\n
By practicing the Counting Sort algorithm through this problem, we learned how to maximize sorting performance. Additionally, we acquired knowledge about basic input\/output handling and array usage in Python. Understanding various sorting algorithms is a critical aspect of programming tests, and approaching problems in this way will be very helpful.<\/p>\n
Additional Problems: Variations and Applications<\/h2>\n
Furthermore, you can try the following variations of the problem:<\/p>\n