Easy loop and condition problem
Intuition\nThe problem seems to ask us to find the sum of squares of elements in the given array num it calculates the sum of squares for every element in nums 
# Intuition\nThe problem seems to ask us to find the sum of squares of elements in the given array num it calculates the sum of squares for every element in nums based on its position rather than its value.\n# Approach\nTo correctly find the sum of squares of elements in the array nums, we need to loop through each element of the array and calculate the square of each element, then add it to the running sum. This can be done using a simple loop.\n# Complexity\nTime complexity: The time complexity of this approach is O(n), where n is the size of the input array nums. We loop through each element once to calculate the sum of squares.\nSpace complexity: The space complexity is O(1) because we are using a constant amount of extra space (only a few variables) regardless of the input array size.\n# Code\n```\nint sumOfSquares(int* nums, int numsSize)\n{   \n   int sum = 0,temp;\n   for(int i = 1; i <= numsSize; i++)\n   {\n       if(numsSize % i == 0)\n       {\n           temp = nums[i-1] * nums[i-1];\n           sum = sum + temp;\n       }\n   }\n   return sum;\n}\n```
['C']