-
Notifications
You must be signed in to change notification settings - Fork 0
Expand file tree
/
Copy pathprogram22.java
More file actions
66 lines (43 loc) · 1.62 KB
/
program22.java
File metadata and controls
66 lines (43 loc) · 1.62 KB
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
/*
2549. Count Distinct Numbers on Board
Easy
166
174
Companies
You are given a positive integer n, that is initially placed on a board. Every day, for 109 days, you perform the following procedure:
For each number x present on the board, find all numbers 1 <= i <= n such that x % i == 1.
Then, place those numbers on the board.
Return the number of distinct integers present on the board after 109 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 is 2.
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.
*/
package LeetCode;
public class program22 {
static int distinctIntegers(int n) {
int cnt=0;
for(int i=1;i<=n;i++){
if( n%i==1){
cnt++;
}
}
return cnt;
}
public static void main(String[] args) {
int n = 5;
System.out.println(distinctIntegers(n));
}
}