determinism's blog

Sum of Increasing Powers of a Number
By determinism, 10 years ago, In English

There are 3 numbers: T, N, M. 1 ≤ T, M ≤ 109, 1 ≤ N ≤ 1018 .

What is asked in the problem is to compute . Obviously, O(N) or O(M) solutions wouldn't work because of 1 second time limit. What can I do?

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10 years ago, # |
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You should in parallel count Tn and , like in binary exponentiation. It's a small exercise to realize how it should work :)

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    10 years ago, # ^ |
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    I just saw this page. By using that formula I can solve it in O(logN), but I couldn't understand what you meant. Can you elaborate a little?

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      10 years ago, # ^ |
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      The main idea is that 1 + T + ... + T2n - 1 = (1 + Tn)·(1 + T + ... + Tn - 1). So, knowing the values of and , we can calculate the desired value. To find sum up to 2n, simply take 1 + T·(1 + T + ... + T2n - 1). So, the algorithm is quite the same as binary exponentiation, but you should keep two values during each step.

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        10 years ago, # ^ |
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        I understand it now, thanks for a great explanation.

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      10 years ago, # ^ |
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      T1 + ... + T2N = (TN + 1)(T1 + ... + TN)

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        10 years ago, # ^ |
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        shouldn't it be (T ^ (N + 1) + 1)( T ^ 1 + ... + T ^ n)? UPD: i was wrong

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          10 years ago, # ^ |
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          nope, notice that there is no T^2n+1 in the summation.

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            10 years ago, # ^ |
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            Sorry, I didn't notice that there T^1, not T^0

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10 years ago, # |
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A more straightforward solution is based on the sum of geometric series : calculate TN + 1 - 1 modulo (T - 1)M with binary exponentiation and divide the remainder by T - 1.

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    10 years ago, # ^ |
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    You could not do it if gcd(T - 1, M) ≠ 1. There is a way to avoid this, counting which powers of prime divisors of M numbers TN + 1 - 1 and T - 1 are divisible by, but it's too complicated.

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      10 years ago, # ^ |
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      I think it will work. Note that I calculate the numerator modulo (T-1)M, not modulo M.

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        10 years ago, # ^ |
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        Oh, sorry, you are right. But if T and M were up to 1018 (maybe this information would be useful for the topicstarter), there would be problems, which we can avoid in the first solution, using binary multiplication.

        And in this solution we should also use binary multiplication, since T·M could be up to 1018. So the complexity is a little worse.

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    10 years ago, # ^ |
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    I also thought about it a moment ago, but then I've realized problem states that T - 1 and M are not necessarily relatively prime.

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      10 years ago, # ^ |
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      Assume that TN + 1 - 1 = k(T - 1)M + r, r is the remainder modulo (T - 1)M. Because T - 1 divides left-hand side, it must divide the right-hand side: r = (T - 1)r'. We obtain , i.e. r' is the remainder modulo M.

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    10 years ago, # ^ |
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    do we need bignum? (T — 1)M can be up to 1e18 — 1e9

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      10 years ago, # ^ |
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      Yeah, Kostroma mentioned above that this would require a 'Russian peasant' modular multiplication.

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        10 years ago, # ^ |
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        thanks for the info

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        10 years ago, # ^ |
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        I read wikipedia's Russian peasant multipication so the overall complexity would be O(lg^2 n) right?

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10 years ago, # |
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slycelote was faster :(

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10 years ago, # |
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f(n) = f(n - 1) * t + 1

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