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p033.java
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p033.java
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/*
* Solution to Project Euler problem 33
* By Nayuki Minase
*
* http://nayuki.eigenstate.org/page/project-euler-solutions
* https://github.com/nayuki/Project-Euler-solutions
*/
public final class p033 implements EulerSolution {
public static void main(String[] args) {
System.out.println(new p033().run());
}
/*
* Consider an arbitrary fraction n/d:
* Let n = 10 * n1 + n0 be the numerator.
* Let d = 10 * d1 + d0 be the denominator.
* As stated in the problem, we need 10 <= n < d < 100.
* We must disregard trivial simplifications where n0 = d0 = 0.
*
* Now, a simplification with n0 = d0 is impossible because:
* n1 / d1 = n / d = (10*n1 + n0) / (10*d1 + n0).
* n1 * (10*d1 + n0) = d1 * (10*n1 + n0).
* 10*n1*d1 + n1*n0 = 10*d1*n1 + d1*n0.
* n1*n0 = d1*n0.
* n1 = d1.
* This implies n = d, which contradicts the fact that n < d.
* Similarly, we cannot have a simplification with n1 = d1 for the same reason.
*
* Therefore we only need to consider the cases where n0 = d1 or n1 = d0.
* In the first case, check that n1/d0 = n/d; in the second case, check that n0/d1 = n/d.
*/
public String run() {
int numer = 1;
int denom = 1;
for (int d = 10; d < 100; d++) {
for (int n = 10; n < d; n++) {
int n0 = n % 10, n1 = n / 10;
int d0 = d % 10, d1 = d / 10;
if (n1 == d0 && n0 * d == n * d1 || n0 == d1 && n1 * d == n * d0) {
numer *= n;
denom *= d;
}
}
}
return Integer.toString(denom / Library.gcd(numer, denom));
}
}