baowilliam's blog

By baowilliam, history, 14 months ago, In English

with $$$(0\leq n\leq 2^{31}-1)$$$ caculate: $$$(2^{2^{n}} + 1)$$$ mod k $$$(1\leq k\leq 10^{6})$$$ i'm doing an exercise on mods for large numbers, i don't know if there is a more efficient solution than using binary exponentiation? Hope to help you, thanks (sorry for my bad english!!!!)

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