Question

1925. Count Square Sum Triples

中文文档

Description

A square triple (a,b,c) is a triple where a, b, and c are integers and a2 + b2 = c2.

Given an integer n, return _the number of square triples such that _1 <= a, b, c <= n.

 

Example 1:

Input: n = 5
Output: 2
Explanation: The square triples are (3,4,5) and (4,3,5).

Example 2:

Input: n = 10
Output: 4
Explanation: The square triples are (3,4,5), (4,3,5), (6,8,10), and (8,6,10).

 

Constraints:

• 1 <= n <= 250

Solutions

Solution 1

class Solution:
  def countTriples(self, n: int) -> int:
    res = 0
    for a in range(1, n + 1):
      for b in range(1, n + 1):
        t = a**2 + b**2
        c = int(sqrt(t))
        if c <= n and c**2 == t:
          res += 1
    return res

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 t = a * a + b * b;
        int c = (int) Math.sqrt(t);
        if (c <= n && c * c == t) {
          ++res;
        }
      }
    }
    return res;
  }
}

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 t = a * a + b * b;
        int c = (int) sqrt(t);
        if (c <= n && c * c == t) {
          ++res;
        }
      }
    }
    return res;
  }
};

func countTriples(n int) int {
  res := 0
  for a := 1; a <= n; a++ {
    for b := 1; b <= n; b++ {
      t := a*a + b*b
      c := int(math.Sqrt(float64(t)))
      if c <= n && c*c == t {
        res++
      }
    }
  }
  return res
}