• 知乎有一個(gè)求助題, 破譯密碼的
• 當(dāng)然了, 根據(jù)普通人定律, 小孩子喜歡用這種方式表白...
• 鏈接在此

https://www.zhihu.com/question/28324597

• 另, 這一看就是凱撒加密...
• 什么是凱撒密碼:

凱撒密碼作為一種最為古老的對(duì)稱加密體制，在古羅馬的時(shí)候都已經(jīng)很流行，他的基本思想是：通過把字母移動(dòng)一定的位數(shù)來實(shí)現(xiàn)加密和解密妹田。明文中的所有字母都在字母表上向后（或向前）按照一個(gè)固定數(shù)目進(jìn)行偏移后被替換成密文狼电。例如飒赃，當(dāng)偏移量是3的時(shí)候喉童，所有的字母A將被替換成D，B變成E滑进，以此類推X將變成A钢颂，Y變成B钞它，Z變成C。由此可見殊鞭，位數(shù)就是凱撒密碼加密和解密的密鑰遭垛。

'VRPHWLPHV L ZDQW WR FKDW ZLWK BRX,EXW L KDYH QR UHDVRQ WR FKDW ZLWK BRX'

def lowercase(s):
    result = ''
    for c in s:
        i = upper.find(c)
        if i == -1:
            # 沒找到, 說明不是大寫字母, 不做處理
            result += c
        else:
            # 找到, 說明它是大寫字母
            result += lower[i]
    return result
def decode(s, shift):
    '''
    :type s: str
    :type shift: int
    :rtype: str
    :param s: 要解密的字符串
    :param shift: 移動(dòng)的位數(shù)
    :return: 解密后的字符串
    '''
    result = ''
    for c in s:
        i = lower.find(c)
        if i == -1:
            result += c
        else:
            # 下面這個(gè)運(yùn)算是為了保證下標(biāo)不為負(fù)數(shù), 反正要對(duì) 26 取模(求余數(shù))
            new_index = (i + 26 - shift) % 26
            code = lower[new_index]
            result += code
    return result

s_lower = lowercase(s)
for i in range(26):
    source = decode(s_lower, i)
    print(source)

運(yùn)行結(jié)果有26個(gè),一目了然,藏在里面

• vrphwlphv l zdqw wr fkdw zlwk brx,exw l kdyh qr uhdvrq wr fkdw zlwk brx
• uqogvkogu k ycpv vq ejcv ykvj aqw,dwv k jcxg pq tgcuqp vq ejcv ykvj aqw
• tpnfujnft j xbou up dibu xjui zpv,cvu j ibwf op sfbtpo up dibu xjui zpv
• sometimes i want to chat with you,but i have no reason to chat with you
• rnldshldr h vzms sn bgzs vhsg xnt,ats h gzud mn qdzrnm sn bgzs vhsg xnt
• qmkcrgkcq g uylr rm afyr ugrf wms,zsr g fytc lm pcyqml rm afyr ugrf wms
• pljbqfjbp f txkq ql zexq tfqe vlr,yrq f exsb kl obxplk ql zexq tfqe vlr
• okiapeiao e swjp pk ydwp sepd ukq,xqp e dwra jk nawokj pk ydwp sepd ukq
• njhzodhzn d rvio oj xcvo rdoc tjp,wpo d cvqz ij mzvnji oj xcvo rdoc tjp
• migyncgym c quhn ni wbun qcnb sio,von c bupy hi lyumih ni wbun qcnb sio
• lhfxmbfxl b ptgm mh vatm pbma rhn,unm b atox gh kxtlhg mh vatm pbma rhn
• kgewlaewk a osfl lg uzsl oalz qgm,tml a zsnw fg jwskgf lg uzsl oalz qgm
• jfdvkzdvj z nrek kf tyrk nzky pfl,slk z yrmv ef ivrjfe kf tyrk nzky pfl
• iecujycui y mqdj je sxqj myjx oek,rkj y xqlu de huqied je sxqj myjx oek
• hdbtixbth x lpci id rwpi lxiw ndj,qji x wpkt cd gtphdc id rwpi lxiw ndj
• gcashwasg w kobh hc qvoh kwhv mci,pih w vojs bc fsogcb hc qvoh kwhv mci
• fbzrgvzrf v jnag gb pung jvgu lbh,ohg v unir ab ernfba gb pung jvgu lbh
• eayqfuyqe u imzf fa otmf iuft kag,ngf u tmhq za dqmeaz fa otmf iuft kag
• dzxpetxpd t hlye ez nsle htes jzf,mfe t slgp yz cpldzy ez nsle htes jzf
• cywodswoc s gkxd dy mrkd gsdr iye,led s rkfo xy bokcyx dy mrkd gsdr iye
• bxvncrvnb r fjwc cx lqjc frcq hxd,kdc r qjen wx anjbxw cx lqjc frcq hxd
• awumbquma q eivb bw kpib eqbp gwc,jcb q pidm vw zmiawv bw kpib eqbp gwc
• zvtlaptlz p dhua av joha dpao fvb,iba p ohcl uv ylhzvu av joha dpao fvb
• yuskzosky o cgtz zu ingz cozn eua,haz o ngbk tu xkgyut zu ingz cozn eua
• xtrjynrjx n bfsy yt hmfy bnym dtz,gzy n mfaj st wjfxts yt hmfy bnym dtz
• wsqixmqiw m aerx xs glex amxl csy,fyx m lezi rs viewsr xs glex amxl csy

知乎鏈接中的各個(gè)答案, 很多大神