写一个函数，输入0打印A,1打印B,...,26打印AA,27打印AB........

wayne

有点长：

1. f = FromCharacterCode[ PadLeft[IntegerDigits[# - (26 (26^Floor[Log[26, 25 #/26 + 1]] - 1))/25, 26], Floor[Log[26, 25 #/26 + 1]] + 1] + 65] &

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反函数：

1. g = FromDigits[ToCharacterCode[#] - 65, 26] + ( 26 (26^(StringLength[#] - 1) - 1))/25 &

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wayne

14# 风云剑 正确！ {10, "K"} {100, "CW"} {1000, "ALM"} {10000, "NTQ"} {100000, "EQXE"} {1000000, "BDWGO"} {10000000, "UVXWK"} {100000000, "HJUNYW"} {1000000000, "CFDGSXM"} {10000000000, "AFIPYQJQ"} {100000000000, "LKRMVRBE"} {1000000000000, "DTMCHRXUO"} {10000000000000, "AUWAGIGNGK"} {100000000000000, "RJVLUNWLUW"} {1000000000000000, "GBDPXGRZXJM"} {10000000000000000, "BRUTMHYHIIZQ"} {100000000000000000, "AAFIWCKSEOVFE"} {1000000000000000000, "JLKTWHMJDBNIO"} {10000000000000000000, "CZTMZYGCWNYMQK"} {100000000000000000000, "ANGWJIRSMASUFQW"}

xbtianlang

const s: String = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'; function AZ(const N: Integer): string; var i: Integer; begin Result:=''; i:=N+1; while i>0 do begin Result:=s[(i-1) mod 26+1]+Result; i:=(i-1) div 26; end; end;

mathe

一个字母的有26个，两个字母的$26^2$个,... 所以字母总数不超过k个的情况共$26+26^2+...+26^k={26^{k+1}-1}/25-1={26^{k+1}}/25-1/25-1=[{26^{k+1}}/25-1]$ 于是对于任意一个数x,如果要找对应的位数，那么就相当于找k是的 ${26^k}/25-1

mathe

比如取x=1000,马上算出对应k=3,由于前面两位数数目是[{26^3}/25-1]=702 然后计算1000-702=298，将298转化为3位26进制数得出(0,11,12),于是结果是ALM

xbtianlang

我的算法是直接递推，当除以26的商=0时结束： $ 1000=((d_3+1)*26+(d_2+1))*26+d_1 $ $ d_1=1000 mod 26 = 12 $ $ d_2= ([1000/26]-1) mod 26 = 11 $ $ d_3= ([([1000/26]-1)/26]-1) mod 26 =0 $ $ 1000=ALM $