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2549. Count Distinct Numbers on Board
Description
You are given a positive integer n
, that is initially placed on a board. Every day, for 10^{9}
days, you perform the following procedure:
 For each number
x
present on the board, find all numbers1 <= i <= n
such thatx % i == 1
.  Then, place those numbers on the board.
Return the number of distinct integers present on the board after 10^{9}
days have elapsed.
Note:
 Once a number is placed on the board, it will remain on it until the end.
%
stands for the modulo operation. For example,14 % 3
is2
.
Example 1:
Input: n = 5 Output: 4 Explanation: Initially, 5 is present on the board. The next day, 2 and 4 will be added since 5 % 2 == 1 and 5 % 4 == 1. After that day, 3 will be added to the board because 4 % 3 == 1. At the end of a billion days, the distinct numbers on the board will be 2, 3, 4, and 5.
Example 2:
Input: n = 3 Output: 2 Explanation: Since 3 % 2 == 1, 2 will be added to the board. After a billion days, the only two distinct numbers on the board are 2 and 3.
Constraints:
1 <= n <= 100
Solutions
Solution 1: Lateral Thinking
Since every operation on the number $n$ on the desktop will also cause the number $n1$ to appear on the desktop, the final numbers on the desktop are $[2,…n]$, that is, $n1$ numbers.
Note that $n$ could be $1$, so it needs to be specially judged.
The time complexity is $O(1)$, and the space complexity is $O(1)$.

class Solution { public int distinctIntegers(int n) { return Math.max(1, n  1); } }

class Solution { public: int distinctIntegers(int n) { return max(1, n  1); } };

class Solution: def distinctIntegers(self, n: int) > int: return max(1, n  1)

func distinctIntegers(n int) int { if n == 1 { return 1 } return n  1 }

function distinctIntegers(n: number): number { return Math.max(1, n  1); }

impl Solution { pub fn distinct_integers(n: i32) > i32 { if n == 1 { return 1; } n  1 } }