c - Why is the run-time efficiency for this code O(n^2)? -

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Sorry, I'm new in bigger analysis. I need some help with it:

I have the following code segment: 

  int sum; {If (n & lt; 1000) sum ++ for (int i = 0; i & lt; n; ++ i); Other zodiac + Fu (n); }   
 The solution key says that it is O (N ^ 2) analysis, but I do not understand why the loop interates only once, so where comes from n ^ 2 is?  
 I'm having trouble with another one:  
 for  (int i = 0; i & lt; n + 100; ++ i) { For (int j = 0; j and lt; i + n; ++ j) yoga = yoga + j; } For (int k = 0; k & lt; n + n + n; ++ k) {c [k] = c [k] + sum; }   
 They come by O (N ^ 3), but I do not even know how it happens. I know that there are 3 loops, but they are not nested, so where does n ^ 3 come from?   
 
 
 1) What could this happen with the implementation of  foo ()  If  foo ()  is  o (n) , this method will be  o (n ^ 2)   
 2) internal Loop goes to  i * n , where  i  goes to  n , thus it is quadratic in  n  The third loop in the two loops  o (n ^ 3)  total is only  o (n)  so analysis can be ignored.

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