Sum of Powers of natural numbers

Правка en1, от UmCaraLegal, 2017-10-03 00:02:17

Be F(N, K) = 1^k + 2^k + 3^k + ... + n^k.

Given N and K we need to calculate F(N, K) (mod 2^32)

INPUT:

1 <= N, K < 2^32

PS: I think about this question for a few days and didn't get success if you have any idea how to solve it, please share it :D

The problem can be found here to submit (statement in portuguese): http://olimpiada.ic.unicamp.br/pratique/programacao/nivels/2013f3p2_somapot

Теги math, advanced math, recurrence

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en1 Английский UmCaraLegal 2017-10-03 00:02:17 454 Initial revision (published)