因为在模意义下需要各种素数。
如果$r \cdot 2^k + 1 $ 是个素数,那么在\(\bmod r \cdot 2^k + 1\)意义下,可以处理 \(2^k\)以内规模的数据。
记录一下 \(a*2^k + 1\)型素数的原根 \(g\)。
\(a*2^k + 1\)\(a\)\(k\)\(g\)\(3\)\(1\)\(1\)\(2\)\(5\)\(1\)\(2\)\(2\)\(17\)\(1\)\(4\)\(3\)\(97\)\(3\)\(5\)\(5\)\(193\)\(3\)\(6\)\(5\)\(257\)\(1\)\(8\)\(3\)\(7681\)\(15\)\(9\)\(17\)\(12289\)\(3\)\(12\)\(11\)\(40961\)\(5\)\(13\)\(3\)\(65537\)\(1\)\(16\)\(3\)\(786433\)\(3\)\(18\)$10 $\(5767169\)\(11\)\(19\)\(3\)\(7340033\)\(7\)$ 20$$ 3$\(23068673\)$ 11 $\(21\)$ 3$\(104857601\)\(25\)\(22\)\(3\)\(167772161\)\(5\)\(25\)$ 3 $\(469762049\)\(7\)$ 26$$ 3$\(998244353(常见)\)$119 $\(23\)$ 3$\(1004535809\)\(479\)$ 21$\(3\)\(1998585857\)\(953\)$ 21 \(|\) 3$\(2013265921\)\(15\)$ 27$\(31\)\(2281701377\)\(17\)$27 \(|\) 3$\(3221225473\)$3 $\(30\)\(5\)\(75161927681\)\(35\)\(31\)$ 3$\(77309411329\)\(9\)$ 33$\(7\)\(206158430209\)\(3\)\(36\)\(22\)\(2061584302081\)\(15\)\(37\)\(7\)\(2748779069441\)$ 5$\(39\)\(3\)\(6597069766657\)\(3\)\(41\)\(5\)\(39582418599937\)\(9\)\(42\)$5 $\(79164837199873\)$ 9$\(43\)$ 5 $\(263882790666241\)\(15\)$ 44$$ 7$\(1231453023109121\)$ 35$$45 $\(3\)\(1337006139375617\)\(19\)\(46\)\(3\)\(3799912185593857\)\(27\)\(47\)\(5\)\(4222124650659841\)\(15\)$ 48 $\(19\)\(7881299347898369\)\(7\)$50 $$6 $\(31525197391593473\)$ 7 $\(52\)\(3\)\(180143985094819841\)$ 5$$55 \(|\) 6$\(1945555039024054273\)$ 27$$ 56 $\(5\)\(4179340454199820289\)$ 29$\(57\)$ 3 $
转载于:https://www.cnblogs.com/LzyRapx/p/9109496.html
相关资源:原根和指数表
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