Add C++ solution for counting the number of valid triples that satisfy the square sum condition in problem 1925.

Author: tkzzzzzz6

leetcode/Study Plan/1925. 统计平方和三元组的数目/1.cpp

@@ -0,0 +1,19 @@
+class Solution{
+    public:
+    int countTriples(int n)
+    {
+        int res = 0;
+        for(int a = 1;a<=n;++a)
+        {
+            for(int b = 1;b<=n;++b)
+            {
+                int c = sqrt(a*a + b*b+1.0);
+                if(c<=n&&c*c == a*a + b*b)
+                {
+                    ++res;
+                }
+            }
+        }
+        return res;
+    }
+};

leetcode/Study Plan/1925. 统计平方和三元组的数目/1.py

@@ -0,0 +1,12 @@
+class Solution:
+    def countTriples(self,n:int)->int:
+        ans = 0
+        for a in range(1,n+1):
+            for b in range(1,a):
+                if a*a+b*b>n*n:
+                    break
+                c2 = a*a+b*b
+                if iSqrt(c2)**2 == c2:
+                    ans+=1
+
+        return ans*2

leetcode/Study Plan/263. 丑数/1.cpp

@@ -0,0 +1,16 @@
+class Solution {
+    public:
+        bool isUgly(int n) {
+            if(n<= 0)
+                return false;
+            while(n%3 == 0)
+            {
+                n /=3;
+            }
+            while(n%5== 0)
+            {
+                n/=5;
+            }
+            return (n & (n-1)) == 0;
+        }
+    };

leetcode/Study Plan/263. 丑数/1.go

@@ -0,0 +1,13 @@
+var factors = []int{2,3,5}
+
+func isUgly(n int) bool {
+	if n<=0{
+		return false
+	}
+	for _,f := range factors{
+		for n%f == 0{
+			n/=f
+		}
+	}
+	return n == 1
+}