Can I use Binary Exponentiation with this problem? (2^2^n + 1) mod k

Правка en4, от baowilliam, 2022-09-26 19:02:58

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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en4 Английский baowilliam 2022-09-26 19:02:58 70 Tiny change: 'with (0\leq n\l' -> 'with $(0\leq n\l'
en3 Английский baowilliam 2022-09-26 18:48:01 10 Tiny change: 'caculate: 2^2^n mod k (1 ' -> 'caculate: (2^2^n + 1) mod k (1 '
en2 Английский baowilliam 2022-09-26 18:47:06 69
en1 Английский baowilliam 2022-09-26 18:44:00 257 Initial revision (published)