-
Notifications
You must be signed in to change notification settings - Fork 0
Expand file tree
/
Copy pathEuler012.java
More file actions
executable file
·72 lines (58 loc) · 1.92 KB
/
Copy pathEuler012.java
File metadata and controls
executable file
·72 lines (58 loc) · 1.92 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
67
68
69
70
71
72
import java.util.Date;
/*
Project Euler Problem 12
========================
The sequence of triangle numbers is generated by adding the natural
numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 =
28. The first ten terms would be:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
Let us list the factors of the first seven triangle numbers:
1: 1
3: 1,3
6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28
We can see that 28 is the first triangle number to have over five
divisors.
What is the value of the first triangle number to have over five hundred
divisors?
*/
public class Euler012 {
public static void main(String[] args) {
Date start, end;
start = new Date();
int divisors = 0;
int n = 1; // the current index of the triangle number used to geneerate
// triangle numbers
int currTriangle = 1;
int o = 0;
while (true) {
divisors = 0;
// since every number has the same number of divisors below the
// square root as above the square root there is no need to count to
// 500. All we need is to find some number with 250 divisors below
// the square root.
o = (int) Math.sqrt(currTriangle);
for (int i = 1; i < o; i++)
divisors = currTriangle % i == 0 ? ++divisors : divisors;
if (divisors >= 250) {
break;
}
currTriangle += ++n;
}
end = new Date();
System.out.printf("The Traingle number is: %d\n", currTriangle);
System.out.println("Execution time = "
+ (end.getTime() - start.getTime()));
}
// method to generate triangle numbers, not called anymore.
public static int triangle(int n) {
int sum = 0;
for (int i = 0; i <= n; i++) {
sum += i;
}
return sum;
}
}