Problem:
Working from left-to-right if no digit is exceeded by the digit to its left it is called an increasing number; for example, 134468.
Similarly if no digit is exceeded by the digit to its right it is called a decreasing number; for example, 66420.
We shall call a positive integer that is neither increasing nor decreasing a "bouncy" number; for example, 155349.
Clearly there cannot be any bouncy numbers below one-hundred, but just over half of the numbers below one-thousand (525) are bouncy. In fact, the least number for which the proportion of bouncy numbers first reaches 50% is 538.
Surprisingly, bouncy numbers become more and more common and by the time we reach 21780 the proportion of bouncy numbers is equal to 90%.
Find the least number for which the proportion of bouncy numbers is exactly 99%.
Similarly if no digit is exceeded by the digit to its right it is called a decreasing number; for example, 66420.
We shall call a positive integer that is neither increasing nor decreasing a "bouncy" number; for example, 155349.
Clearly there cannot be any bouncy numbers below one-hundred, but just over half of the numbers below one-thousand (525) are bouncy. In fact, the least number for which the proportion of bouncy numbers first reaches 50% is 538.
Surprisingly, bouncy numbers become more and more common and by the time we reach 21780 the proportion of bouncy numbers is equal to 90%.
Find the least number for which the proportion of bouncy numbers is exactly 99%.
Solution:
76576500
Code:
The solution may include methods that will be found here: Library.java .
public interface EulerSolution{
public String run();
}
/*
* Solution to Project Euler problem 12
* By Nayuki Minase
*
* http://nayuki.eigenstate.org/page/project-euler-solutions
* https://github.com/nayuki/Project-Euler-solutions
*/
public final class p012 implements EulerSolution {
public static void main(String[] args) {
System.out.println(new p012().run());
}
public String run() {
int num = 0;
for (int i = 1; ; i++) {
num += i; // num is triangle number i
if (countDivisors(num) > 500)
return Integer.toString(num);
}
}
private static int countDivisors(int n) {
int count = 0;
int end = Library.sqrt(n);
for (int i = 1; i < end; i++) {
if (n % i == 0)
count += 2;
}
if (end * end == n) // Perfect square
count++;
return count;
}
}
No comments :
Post a Comment