1201. Ugly Number I I I

1201. Ugly Number III #

题目 #

Write a program to find the n-th ugly number.

Ugly numbers are positive integers which are divisible by a or b or c.

Example 1:

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

Example 2:

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

Example 3:

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

Example 4:

Input: n = 1000000000, a = 2, b = 217983653, c = 336916467
Output: 1999999984

Constraints:

  • 1 <= n, a, b, c <= 10^9
  • 1 <= a * b * c <= 10^18
  • It’s guaranteed that the result will be in range [1, 2 * 10^9]

题目大意 #

请你帮忙设计一个程序,用来找出第 n 个丑数。丑数是可以被 a 或 b 或 c 整除的 正整数。

提示:

  • 1 <= n, a, b, c <= 10^9
  • 1 <= a * b * c <= 10^18
  • 本题结果在 [1, 2 * 10^9] 的范围内

解题思路 #

  • 给出 4 个数字,a,b,c,n。要求输出可以整除 a 或者整除 b 或者整除 c 的第 n 个数。
  • 这一题限定了解的范围, [1, 2 * 10^9],所以直接二分搜索来求解。逐步二分逼近 low 值,直到找到能满足条件的 low 的最小值,即为最终答案。
  • 这一题的关键在如何判断一个数是第几个数。一个数能整除 a,能整除 b,能整除 c,那么它应该是第 num/a + num/b + num/c - num/ab - num/bc - num/ac + num/abc 个数。这个就是韦恩图。需要注意的是,求 abbcacabc 的时候需要再除以各自的最大公约数 gcd()

代码 #


package leetcode

func nthUglyNumber(n int, a int, b int, c int) int {
	low, high := int64(0), int64(2*1e9)
	for low < high {
		mid := low + (high-low)>>1
		if calNthCount(mid, int64(a), int64(b), int64(c)) < int64(n) {
			low = mid + 1
		} else {
			high = mid
		}
	}
	return int(low)
}

func calNthCount(num, a, b, c int64) int64 {
	ab, bc, ac := a*b/gcd(a, b), b*c/gcd(b, c), a*c/gcd(a, c)
	abc := a * bc / gcd(a, bc)
	return num/a + num/b + num/c - num/ab - num/bc - num/ac + num/abc
}

func gcd(a, b int64) int64 {
	for b != 0 {
		a, b = b, a%b
	}
	return a
}


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