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shellhub 2020-10-06 14:58:38 +08:00
parent 064d941722
commit 8ad87ce9d3
1 changed files with 36 additions and 33 deletions
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69
ProjectEuler/Problem12.java
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@ -1,24 +1,34 @@
/*
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:
package ProjectEuler;
/**
* 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:
* <p>
* 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
* <p>
* Let us list the factors of the first seven triangle numbers:
* <p>
* 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.
* <p>
* What is the value of the first triangle number to have over five hundred divisors?
* <p>
* link: https://projecteuler.net/problem=12
*/
public class Problem12 {
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 Problem_12 {
/**
* Driver Code
*/
public static void main(String[] args) {
assert solution1(500) == 76576500;
}
/* returns the nth triangle number; that is, the sum of all the natural numbers less than, or equal to, n */
public static int triangleNumber(int n) {
@ -28,36 +38,29 @@ public class Problem_12 {
return sum;
}
public static void main(String[] args) {
long start = System.currentTimeMillis(); // start the stopwatch
public static int solution1(int number) {
int j = 0; // j represents the jth triangle number
int n = 0; // n represents the triangle number corresponding to j
int numberOfDivisors = 0; // number of divisors for triangle number n
while (numberOfDivisors <= 500) {
while (numberOfDivisors <= number) {
// resets numberOfDivisors because it's now checking a new triangle number
// and also sets n to be the next triangle number
numberOfDivisors = 0;
j++;
n = triangleNumber(j);
// for every number from 1 to the square root of this triangle number,
// count the number of divisors
for (int i = 1; i <= Math.sqrt(n); i++)
if (n % i == 0)
numberOfDivisors++;
// 1 to the square root of the number holds exactly half of the divisors
// so multiply it by 2 to include the other corresponding half
numberOfDivisors *= 2;
}
long finish = System.currentTimeMillis(); // stop the stopwatch
System.out.println(n);
System.out.println("Time taken: " + (finish - start) + " milliseconds");
return n;
}
}