leetcode 878. Nth Magical Number

景行cmy

于 2019-04-12 16:37:39 发布

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本文介绍了一种高效算法，用于找到第N个能被A或B整除的正整数，即所谓的魔法数。通过使用最小公倍数和二分查找，算法能在大规模数据下快速求解，返回结果模10^9+7。

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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 <= N <= 10^9
2 <= A <= 40000
2 <= B <= 40000

class Solution {
public:
  int nthMagicalNumber(int N, int A, int B) 
  {
      long kmod = 1e9 +7;
      long left = 0;
      long right = 1e9*4e5+1;
      long mid=0;
      int d = lcm(A,B);
      while(left<right)
      {
          mid = left + (right-left)/2;
          if((mid/A+mid/B-mid/d)<N) left = mid+1;
          else right = mid;
      }
      return left%kmod;
  }

};