Hi, 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 about my bad english :()
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Can I use Binary Exponentiation with this problem? 2^2^n mod k
Hi, 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 about my bad english :()
Rev. | Lang. | By | When | Δ | Comment | |
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en4 |
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baowilliam | 2022-09-26 19:02:58 | 70 | Tiny change: 'with (0\leq n\l' -> 'with $(0\leq n\l' | |
en3 |
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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 |
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baowilliam | 2022-09-26 18:47:06 | 69 | ||
en1 |
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baowilliam | 2022-09-26 18:44:00 | 257 | Initial revision (published) |
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