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feat: add solutions to lc problem: No.1201 (doocs#856)
No.1201.Ugly Number III
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‎solution/1200-1299/1201.Ugly Number III/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：二分搜索**
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根据题目提示结果在 [1, 2 * 10<sup>9</sup>] 的闭区间上，所以定义二分搜索的左边界 left=1，右边界 right=2e9。此时我们只需要在 [left,right] 的闭区间内找到一个最小的数 num，使其满足 [1,num] 内的丑数总数等于 n，则 num 就是第 n 个丑数。计算在 [1,num] 的范围内丑数的数目，即可以被 a、b 或 c 任意一个数整除的数的总数，其方法如下：
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`f(num, a, b, c) = num/a + num/b + num/c - a⋂b - a⋂c - b⋂c + a⋂b⋂c`
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- num/a 表示在 [1,num] 内可以整除 a 的数目，num/b 表示在 [1,num] 内可以整除 b 的数目，num/c 表示在 [1,num] 内可以整除 c 的数目。
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- a⋂b 表示在 [1,num] 内可以同时整除 a 和 b 的数目，a⋂c 表示在 [1,num] 内可以同时整除 a 和 c 的数，b⋂c 表示在 [1,num] 内可以同时整除 b 和 c 的数。
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- a⋂b⋂c 表示在 [1,num] 内可以同时整除 a、b 和 c 的数。
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- a⋂b = num/least_common_multiple(a, b)，其他情况依次类推。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def f(self, num: int, a: int, b: int, c: int) -> int:
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return num // a + num // b + num // c - num // math.lcm(a, b) - num // math.lcm(a, c) - num // math.lcm(b, c) \
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+ num // math.lcm(a, b, c)
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def nthUglyNumber(self, n: int, a: int, b: int, c: int) -> int:
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left, right = 1, int(2e9)
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while left <= right:
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mid = left + (right - left) // 2
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if self.f(mid, a, b, c) < n:
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left = mid + 1
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else:
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right = mid - 1
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return left
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```
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### **Java**
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### **Go**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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```go
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func nthUglyNumber(n int, a int, b int, c int) int {
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left, right := 1, int(2e9)
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for left <= right {
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mid := left + (right-left)/2
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if f(mid, a, b, c) < n {
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left = mid + 1
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} else {
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right = mid - 1
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}
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}
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return left
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}
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func f(num int, a int, b int, c int) int {
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return num/a + num/b + num/c - num/lcm(a, b) - num/lcm(a, c) - num/lcm(b, c) + num/lcm(lcm(a, b), c)
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}
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// Least common multiple
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func lcm(a, b int) int {
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// Greatest common divisor
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gcd := func(x, y int) int {
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for y != 0 {
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if x < y {
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x, y = y, x
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}
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x, y = y, x%y
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}
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return x
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}
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return a * b / gcd(a, b)
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}
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```
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### **C++**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```cpp
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class Solution {
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public:
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long gcd(long x, long y) {
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while (y != 0) {
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if (x < y)
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swap(x, y);
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long tmp = x % y;
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x = y;
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y = tmp;
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}
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return x;
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}
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long lcm(long x, long y) { return x * y / gcd(x, y); }
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long f(int num, int a, int b, int c) {
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long sumabc = long(num / a) + num / b + num / c;
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long intersections = long(num / lcm(a, b)) + num / lcm(a, c) + num / lcm(b, c) - num / lcm(lcm(a, b), c);
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return sumabc - intersections;
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}
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int nthUglyNumber(int n, int a, int b, int c) {
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int left = 1, right = int(2e9);
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while (left <= right) {
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int mid = left + (right - left) / 2;
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if (f(mid, a, b, c) < n) {
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left = mid + 1;
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} else {
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right = mid - 1;
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}
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}
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return left;
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}
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};
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```
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### **...**

‎solution/1200-1299/1201.Ugly Number III/README_EN.md

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### **Python3**
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```python
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class Solution:
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def f(self, num: int, a: int, b: int, c: int) -> int:
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return num // a + num // b + num // c - num // math.lcm(a, b) - num // math.lcm(a, c) - num // math.lcm(b, c) \
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+ num // math.lcm(a, b, c)
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def nthUglyNumber(self, n: int, a: int, b: int, c: int) -> int:
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left, right = 1, int(2e9)
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while left <= right:
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mid = left + (right - left) // 2
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if self.f(mid, a, b, c) < n:
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left = mid + 1
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else:
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right = mid - 1
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return left
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```
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### **Java**
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```java
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### **Go**
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```go
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func nthUglyNumber(n int, a int, b int, c int) int {
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left, right := 1, int(2e9)
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for left <= right {
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mid := left + (right-left)/2
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if f(mid, a, b, c) < n {
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left = mid + 1
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} else {
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right = mid - 1
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}
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}
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return left
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}
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func f(num int, a int, b int, c int) int {
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return num/a + num/b + num/c - num/lcm(a, b) - num/lcm(a, c) - num/lcm(b, c) + num/lcm(lcm(a, b), c)
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}
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// Least common multiple
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func lcm(a, b int) int {
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// Greatest common divisor
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gcd := func(x, y int) int {
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for y != 0 {
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if x < y {
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x, y = y, x
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}
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x, y = y, x%y
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}
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return x
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}
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return a * b / gcd(a, b)
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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long gcd(long x, long y) {
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while (y != 0) {
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if (x < y)
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swap(x, y);
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long tmp = x % y;
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x = y;
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y = tmp;
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}
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return x;
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}
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long lcm(long x, long y) { return x * y / gcd(x, y); }
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long f(int num, int a, int b, int c) {
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long sumabc = long(num / a) + num / b + num / c;
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long intersections = long(num / lcm(a, b)) + num / lcm(a, c) + num / lcm(b, c) - num / lcm(lcm(a, b), c);
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return sumabc - intersections;
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}
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int nthUglyNumber(int n, int a, int b, int c) {
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int left = 1, right = int(2e9);
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while (left <= right) {
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int mid = left + (right - left) / 2;
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if (f(mid, a, b, c) < n) {
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left = mid + 1;
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} else {
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right = mid - 1;
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}
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}
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return left;
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}
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};
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```
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### **...**

‎solution/1200-1299/1201.Ugly Number III/Solution.cpp

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class Solution {
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public:
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long gcd(long x, long y) {
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while (y != 0) {
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if (x < y)
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swap(x, y);
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long tmp = x % y;
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x = y;
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y = tmp;
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}
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return x;
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}
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long lcm(long x, long y) { return x * y / gcd(x, y); }
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long f(int num, int a, int b, int c) {
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long sumabc = long(num / a) + num / b + num / c;
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long intersections = long(num / lcm(a, b)) + num / lcm(a, c) + num / lcm(b, c) - num / lcm(lcm(a, b), c);
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return sumabc - intersections;
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}
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int nthUglyNumber(int n, int a, int b, int c) {
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int left = 1, right = int(2e9);
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while (left <= right) {
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int mid = left + (right - left) / 2;
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if (f(mid, a, b, c) < n) {
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left = mid + 1;
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} else {
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right = mid - 1;
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}
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}
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return left;
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}
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};

‎solution/1200-1299/1201.Ugly Number III/Solution.go

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func nthUglyNumber(n int, a int, b int, c int) int {
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left, right := 1, int(2e9)
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for left <= right {
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mid := left + (right-left)/2
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if f(mid, a, b, c) < n {
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left = mid + 1
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} else {
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right = mid - 1
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}
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}
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return left
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}
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func f(num int, a int, b int, c int) int {
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return num/a + num/b + num/c - num/lcm(a, b) - num/lcm(a, c) - num/lcm(b, c) + num/lcm(lcm(a, b), c)
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}
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// Least common multiple
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func lcm(a, b int) int {
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// Greatest common divisor
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gcd := func(x, y int) int {
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for y != 0 {
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if x < y {
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x, y = y, x
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}
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x, y = y, x%y
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}
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return x
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}
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return a * b / gcd(a, b)
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}

‎solution/1200-1299/1201.Ugly Number III/Solution.py

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class Solution:
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def f(self, num: int, a: int, b: int, c: int) -> int:
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return num // a + num // b + num // c - num // math.lcm(a, b) - num // math.lcm(a, c) - num // math.lcm(b, c) \
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+ num // math.lcm(a, b, c)
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def nthUglyNumber(self, n: int, a: int, b: int, c: int) -> int:
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left, right = 1, int(2e9)
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while left <= right:
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mid = left + (right - left) // 2
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if self.f(mid, a, b, c) < n:
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left = mid + 1
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else:
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right = mid - 1
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return left

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