Ugly Number III
<p>An <strong>ugly number</strong> is a positive integer that is divisible by <code>a</code>, <code>b</code>, or <code>c</code>.</p>

<p>Given four integers <code>n</code>, <code>a</code>, <code>b</code>, and <code>c</code>, return the <code>n<sup>th</sup></code> <strong>ugly number</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 3, a = 2, b = 3, c = 5
<strong>Output:</strong> 4
<strong>Explanation:</strong> The ugly numbers are 2, 3, 4, 5, 6, 8, 9, 10... The 3<sup>rd</sup> is 4.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 4, a = 2, b = 3, c = 4
<strong>Output:</strong> 6
<strong>Explanation:</strong> The ugly numbers are 2, 3, 4, 6, 8, 9, 10, 12... The 4<sup>th</sup> is 6.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 5, a = 2, b = 11, c = 13
<strong>Output:</strong> 10
<strong>Explanation:</strong> The ugly numbers are 2, 4, 6, 8, 10, 11, 12, 13... The 5<sup>th</sup> is 10.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n, a, b, c &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= a * b * c &lt;= 10<sup>18</sup></code></li>
	<li>It is guaranteed that the result will be in range <code>[1, 2 * 10<sup>9</sup>]</code>.</li>
</ul>
Math; Binary Search; Combinatorics; Number Theory
class Solution {
    public int nthUglyNumber(int n, int a, int b, int c) {
        long ab = lcm(a, b);
        long bc = lcm(b, c);
        long ac = lcm(a, c);
        long abc = lcm(ab, c);
        long l = 1, r = 2000000000;
        while (l < r) {
            long mid = (l + r) >> 1;
            if (mid / a + mid / b + mid / c - mid / ab - mid / bc - mid / ac + mid / abc >= n) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return (int) l;
    }

    private long gcd(long a, long b) {
        return b == 0 ? a : gcd(b, a % b);
    }

    private long lcm(long a, long b) {
        return a * b / gcd(a, b);
    }
}