「BZOJ1965」[Ahoi2005] SHUFFLE 洗牌

2014年12月30日4,1071

\[x*(2^m)\equiv l(mod~n+1)\]x在mod n+1下逆元是n/2+1

所以移项得\[x\equiv(n/2+1)^m*l(mod~n+1)\]

 

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kadfin
kadfin

应该是 2在mod n+1下逆元是n/2+1