Sum of Square Numbers in C++

Suppose we have a non-negative integer c, we have to decide whether there're two integers a and b such that it satisfies a^2 + b^2 = c.

So, if the input is like 61, then the output will be True, as 61 = 5^2 + 6^2.

To solve this, we will follow these steps −

• Define a function isPerfect(), this will take x,

• sr := square root of x

• return true when (sr - floor of sr) is 0

• From the main method do the following,

• if c is same as 0, then −

  • return true

• for initialize i := 0, when i < the ceiling of square root of c, update (increase i by 1), do −

  • b := c - i * i

  • if isPerfect(b) is true, then −

    • return true

• return false

Example 

Let us see the following implementation to get a better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
   bool isPerfect(int x){
      long double sr = sqrt(x);
      return ((sr - floor(sr)) == 0);
   }
   bool judgeSquareSum(int c) {
      if (c == 0)
         return true;
      int b;
      for (int i = 0; i < ceil(sqrt(c)); i++) {
         b = c - i * i;
         if (isPerfect(b))
            return true;
      }
      return false;
   }
};
main(){
   Solution ob;
   cout << (ob.judgeSquareSum(61));
}

Input

61

Output

1