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 #

• 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 `ab``bc``ac``abc` 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
}

``````

Sep 6, 2020