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81 lines
1.6 KiB
Markdown
Executable File
81 lines
1.6 KiB
Markdown
Executable File
# [878. Nth Magical Number](https://leetcode.com/problems/nth-magical-number/)
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## 题目
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A positive integer is *magical* if it is divisible by either A or B.
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Return the N-th magical number. Since the answer may be very large, **return it modulo** `10^9 + 7`.
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**Example 1**:
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Input: N = 1, A = 2, B = 3
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Output: 2
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**Example 2**:
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Input: N = 4, A = 2, B = 3
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Output: 6
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**Example 3**:
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Input: N = 5, A = 2, B = 4
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Output: 10
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**Example 4**:
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Input: N = 3, A = 6, B = 4
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Output: 8
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**Note**:
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1. `1 <= N <= 10^9`
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2. `2 <= A <= 40000`
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3. `2 <= B <= 40000`
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## 题目大意
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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。
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提示：
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1. 1 <= N <= 10^9
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2. 2 <= A <= 40000
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3. 2 <= B <= 40000
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## 解题思路
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- 给出 3 个数字，a，b，n。要求输出可以整除 a 或者整除 b 的第 n 个数。
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- 这一题是第 1201 题的缩水版，代码和解题思路也基本不变，这一题的二分搜索的区间是 `[min(A, B)，N * min(A, B)] = [2, 10 ^ 14]`。其他代码和第 1201 题一致，思路见第 1201 题。
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## 代码
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```go
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package leetcode
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func nthMagicalNumber(N int, A int, B int) int {
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low, high := int64(0), int64(1*1e14)
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for low < high {
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mid := low + (high-low)>>1
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if calNthMagicalCount(mid, int64(A), int64(B)) < int64(N) {
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low = mid + 1
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} else {
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high = mid
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}
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}
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return int(low) % 1000000007
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}
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func calNthMagicalCount(num, a, b int64) int64 {
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ab := a * b / gcd(a, b)
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return num/a + num/b - num/ab
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}
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``` |