添加 problem 1201

Algorithms/1201. Ugly Number III/1201. Ugly Number III.go

@@ -0,0 +1,27 @@
+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
+}

Algorithms/1201. Ugly Number III/1201. Ugly Number III_test.go

@@ -0,0 +1,60 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1201 struct {
+	para1201
+	ans1201
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1201 struct {
+	n int
+	a int
+	b int
+	c int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1201 struct {
+	one int
+}
+
+func Test_Problem1201(t *testing.T) {
+
+	qs := []question1201{
+
+		question1201{
+			para1201{3, 2, 3, 5},
+			ans1201{4},
+		},
+
+		question1201{
+			para1201{4, 2, 3, 4},
+			ans1201{6},
+		},
+
+		question1201{
+			para1201{5, 2, 11, 13},
+			ans1201{10},
+		},
+
+		question1201{
+			para1201{1000000000, 2, 217983653, 336916467},
+			ans1201{1999999984},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1201------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1201, q.para1201
+		fmt.Printf("【input】:%v       【output】:%v\n", p, nthUglyNumber(p.n, p.a, p.b, p.c))
+	}
+	fmt.Printf("\n\n\n")
+}

Algorithms/1201. Ugly Number III/README.md

@@ -0,0 +1,57 @@
+# [1201. Ugly Number III](https://leetcode.com/problems/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 `ab`、`bc`、`ac`、`abc` 的时候需要再除以各自的最大公约数 `gcd()`。