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Without resorting to asymptotic notation, is tedious step counting the only way to get the time complexity of an algorithm? And without step count of each line of code can we arrive at a big-O representation of any program?
Details: trying to find out the complexity of several numerical analysis algorithms to decide which will be best suited for solving a particular problem. E.g. - from among Regula-Falsi or Newton-Rhapson method for solving eqns, intention is to evaluate the exact complexity of each method and then decide (putting value of 'n' or whatever arguments there are) which method is less complex.
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If your algorithm runs in a time proportional to the logarithm of the input data size, that is log ( n ) \log(n) log(n), then you have O ( log ( n ) ) \mathcal{O}(\log(n)) O(log(n)) complexity. This type of complexity is usually present in algorithms that somehow divide the input size.
The time complexity of a loop is equal to the number of times the innermost statement is to be executed. On the first iteration of i=0, the inner loop executes 0 times. On the first iteration of i=1, the inner loop executes 1 times. On the first iteration of i=n-1, the inner loop executes n-1 times.
The only way --- not the "easy" or hard way but the only reasonable way --- to find the exact complexity of a complicated algorithm is to profile it. A modern implementation of an algorithm has a complex interaction with numerical libraries and with the CPU and its floating point unit. For instance in-cache memory access is much faster than out-of-cache memory access, and on top of that there may be more than one level of cache. Counting steps is really much more suitable to the asymptotic complexity that you say isn't enough for your purpose.
But, if you did want to count steps automatically, there are also ways to do that. You can add a counter increment command (like "bloof++;" in C) to every line of code, and then display the value at the end.
You should also know about the more refined time complexity expression, f(n)*(1+o(1)), that is also useful for analytical calculations. For instance n^2+2*n+7 simplifies to n^2*(1+o(1)). If the constant factor is what bothers you about usual asymptotic notation O(f(n)), this refinement is a way to keep track of it and still throw out negligible terms.
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