feat: update python3 solution to lc problem: No.2992 (#4091)

Author: fxrcode

solution/2900-2999/2992.Number of Self-Divisible Permutations/README.md

@@ -108,7 +108,7 @@ class Solution:
                 return 1
             ans = 0
             for j in range(1, n + 1):
-                if (mask >> j & 1) == 0 and (i % j == 0 or j % i == 0):
+                if (mask >> j & 1) == 0 and gcd(i, j) == 1:
                     ans += dfs(mask | 1 << j)
             return ans
 
@@ -267,7 +267,7 @@ class Solution:
         for mask in range(1 << n):
             i = mask.bit_count()
             for j in range(1, n + 1):
-                if (mask >> (j - 1) & 1) == 1 and (i % j == 0 or j % i == 0):
+                if (mask >> (j - 1) & 1) == 1 and gcd(i, j) == 1:
                     f[mask] += f[mask ^ (1 << (j - 1))]
         return f[-1]
 ```

solution/2900-2999/2992.Number of Self-Divisible Permutations/README_EN.md

@@ -106,7 +106,7 @@ class Solution:
                 return 1
             ans = 0
             for j in range(1, n + 1):
-                if (mask >> j & 1) == 0 and (i % j == 0 or j % i == 0):
+                if (mask >> j & 1) == 0 and gcd(i, j) == 1:
                     ans += dfs(mask | 1 << j)
             return ans
 
@@ -265,7 +265,7 @@ class Solution:
         for mask in range(1 << n):
             i = mask.bit_count()
             for j in range(1, n + 1):
-                if (mask >> (j - 1) & 1) == 1 and (i % j == 0 or j % i == 0):
+                if (mask >> (j - 1) & 1) == 1 and gcd(i, j) == 1:
                     f[mask] += f[mask ^ (1 << (j - 1))]
         return f[-1]
 ```

solution/2900-2999/2992.Number of Self-Divisible Permutations/Solution.py

@@ -7,7 +7,7 @@ def dfs(mask: int) -> int:
                 return 1
             ans = 0
             for j in range(1, n + 1):
-                if (mask >> j & 1) == 0 and (i % j == 0 or j % i == 0):
+                if (mask >> j & 1) == 0 and gcd(i, j) == 1:
                     ans += dfs(mask | 1 << j)
             return ans
 

solution/2900-2999/2992.Number of Self-Divisible Permutations/Solution2.py

@@ -5,6 +5,6 @@ def selfDivisiblePermutationCount(self, n: int) -> int:
         for mask in range(1 << n):
             i = mask.bit_count()
             for j in range(1, n + 1):
-                if (mask >> (j - 1) & 1) == 1 and (i % j == 0 or j % i == 0):
+                if (mask >> (j - 1) & 1) == 1 and gcd(i, j) == 1:
                     f[mask] += f[mask ^ (1 << (j - 1))]
         return f[-1]