Time complexity double loops

Find the number of times f() is called as a function of the input size n.

a. for (i = n; i > 1; i /= 2) 
 		f()
a. O(log n)

b. for (i = 0; i * i < n; i++) 
		f()

b. O(sqrt n)
 
c. for (i = 0; i < n; i++)
	  	for (j = 0; j < 256; j++)
		  f()
	
c. O(n)

  
d. for (i = 1; i <= n * n - 10; i++) 
		for (j = 1; j <= i; j ++)
			f()

d. O(n**4)

e.  for (i = 1; i <= n; i++)
		for (j = 1; j <= i; j++)
			for (k = 1; k <= j; k++) 
				f()
e. O(n**3)

f. for (i = 1; i <= n; i++)
		for (j = 1; j < i; j *= 2)
			f()

f. O(logn!) = O(nlgn)

g. for (i = 1; i <= n; i++)
		for (j = 1; j <= n; j += i)
			f()
g. O(n*lnn)

h. void compute(int n)
	  if (n == 0) return;
	  for (int i = 0; i < n; i++) 
	  	  f()
	  compute(n/2) 
	  compute(n/2)
h. O(n*logn)

i. for (int i = 8; i < n + 7; i += 2) { f(); }
f() is called (n + 7 − 8)/2 times, which is in the order of n.

j. for (int i = 8; i < n + 7; i *= 3) { f(); }
8 * 3**i >= n + 7 => f is called log3 (n+7)/8 times

j. for(int i = 0; i < n; ++i) {
	for(int j = 0; j < n; ++j)
		f();
	for (int j = 1; j < n; j *= 2)
		f();
}
n*(n + logn)