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[Swift]LeetCode878. 第 N 个神奇数字 | Nth Magical Number

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A positive integer is magical if it is divisible by either A or B.

Return the N-th magical number.  Since the answer may be very large, return it modulo 10^9 + 7

Example 1:

Input: N = 1, A = 2, B = 3
Output: 2

Example 2:

Input: N = 4, A = 2, B = 3
Output: 6

Example 3:

Input: N = 5, A = 2, B = 4
Output: 10

Example 4:

Input: N = 3, A = 6, B = 4
Output: 8 

Note:

  1. 1 <= N <= 10^9
  2. 2 <= A <= 40000
  3. 2 <= B <= 40000

如果正整数可以被 A 或 B 整除,那么它是神奇的。

返回第 N 个神奇数字。由于答案可能非常大,返回它模 10^9 + 7 的结果。 

示例 1:

输入:N = 1, A = 2, B = 3
输出:2

示例 2:

输入:N = 4, A = 2, B = 3
输出:6

示例 3:

输入:N = 5, A = 2, B = 4
输出:10

示例 4:

输入:N = 3, A = 6, B = 4
输出:8 

提示:

  1. 1 <= N <= 10^9
  2. 2 <= A <= 40000
  3. 2 <= B <= 40000

Runtime: 4 ms
Memory Usage: 18.4 MB
 1 class Solution {
 2     func nthMagicalNumber(_ N: Int, _ A: Int, _ B: Int) -> Int {
 3         var a:Int = A
 4         var b:Int = B
 5         var tmp:Int = 0
 6         var l:Int = 2
 7         var r:Int = Int(1e14)
 8         var mod:Int = Int(1e9 + 7)
 9         while (b > 0)
10          {
11             tmp = a
12             a = b
13             b = tmp % b
14         }
15         while (l < r)
16         {
17             var m:Int = (l + r) / 2;
18             if m / A + m / B - m / (A * B / a) < N
19             {
20                 l = m + 1
21             }
22             else
23             {
24                 r = m
25             }
26         }
27         return Int(l % mod)
28     }
29 }

 

posted @ 2019-03-26 16:33  为敢技术  阅读(296)  评论(0编辑  收藏  举报