答案：（a）已知F解密为w 频数统计为：
A = 5 B = 0 C = 37 D = 8 E = 12 F = 9 G = 23 H = 5 I = 15 J = 7 K = 17 L = 7 M = 5 N = 13 O = 10 P = 6 Q = 1 R = 0 S = 20 T = 0 U = 14 V = 0 W = 5 X = 7 Y = 15 Z = 13 F解密后，为w
EMGLOSUDCGDNCUSWYSwHNSwCYKDPUMLWGYICOXYSIPJCKQPKUGKMGOLICGINCGACKSNISACYKZSCKXECJCKSHYSXCGOIDPKZCNKSHICGIWYGKKGOLDSILKGOIUSIGLEDSPWZUGwZCCNDGYYSwUSZCNXEOJNCGYEOWEUPXEZGACGNwGLKNSACIGOIYCKXCJUCIUZCwZCCNDGYYSwEUEKUZCSOCwZCCNCIACZEJNCSHwZEJZEGMXCYHCJUMGKUCY
有频数，大胆猜测C解密为e 。有
EMGLOSUDeGDNeUSWYSwHNSweYKDPUMLWGYIeOXYSIPJeKQPKUGKMGOLIeGINeGAeKSNISAeYKZSeKXEeJeKSHYSXeGOIDPKZeNKSHIeGIWYGKKGOLDSILKGOIUSIGLEDSPWZUGwZeeNDGYYSwUSZeNXEOJNeGYEOWEUPXEZGAeGNwGLKNSAeIGOIYeKXeJUeIUZewZeeNDGYYSwEUEKUZeSOewZeeNeIAeZEJNeSHwZEJZEGMXeYHeJUMGKUeY
在和e的双字母组合中eG出现了7次，Ze出现了7次，eK出现了5次，Ae出现了5次，eY出现了4次，eI出现了4次而且Ie出现了3次，Ne出现了4次，且Ge未出现一次，而G出现23次较高，故猜测I 解密为{r,s,t}中一个，S出现概率较高与e结合较少，猜测S为o，有
EMGLOoUDeGDNeUoWYowHNoweYKDPUMLWGYIeOXYoIPJeKQPKUGKMGOLIeGINeGAeKoNIoAeYKZoeKXEeJeKoHYoXeGOIDPKZeNKoHIeGIWYGKKGOLDoILKGOIUoIGLEDoPWZUGwZeeNDGYYowUoZeNXEOJNeGYEOWEUPXEZGAeGNwGLKNoAeIGOIYeKXeJUeIUZewZeeNDGYYowEUEKUZeoOewZeeNeIAeZEJNeoHwZEJZEGMXeYHeJUMGKUeY
已知eG,eK出现较多，而Ge，Ke未出现，所以G，K中有一个为a , Ze出现较多，且wZ出现较多，猜测Z为h,有：
EMGLOoUDeGDNeUoWYowHNoweYKDPUMLWGYIeOXYoIPJeKQPKUGKMGOLIeGINeGAeKoNIoAeYKhoeKXEeJeKoHYoXeGOIDPKheNKoHIeGIWYGKKGOLDoILKGOIUoIGLEDoPWhUGwheeNDGYYowUoheNXEOJNeGYEOWEUPXEhGAeGNwGLKNoAeIGOIYeKXeJUeIUhewheeNDGYYowEUEKUheoOewheeNeIAehEJNeoHwhEJhEGMXeYHeJUMGKUeY
whee为开头出现了3次，后面均为N，猜测whee为单词开头，N应为{l,z}中一个，wheeNDGYYow出现两次，恰有一单词吻合，猜测其为wheelbarrow，则N解密为l D解密为b，G解密为a，Y解密为r。有
EMaLOoUbeableUoWrowHlowerKbPUMLWarIeOXroIPJeKQPKUaKMaOLIeaIleaAeKolIoAerKhoeKXEeJeKoHroXeaOIbPKhelKoHIeaIWraKKaOLboILKaOIUoIaLEboPWhUawheelbarrowUohelXEOJlearEOWEUPXEhaAealwaLKloAeIaOIreKXeJUeIUhewheelbarrowEUEKUheoOewheeleIAehEJleoHwhEJhEaMXerHeJUMaKUer
已知Uo和Ko出现了3次，由Uh和Kh出现了3次，U，K应该为s和t的顺序 由Uohel知Uo应为一单词，U应该为t，K为s。有
EMaLOotbeabletoWrowHlowersbPtMLWarIeOXroIPJesQPstasMaOLIeaIleaAesolIoAershoesXEeJesoHroXeaOIbPshelsoHIeaIWrassaOLboILsaOItoIaLEboPWhtawheelbarrowtohel
XEOJlearEOWEtPXEhaAealwaLsloAeIaOIresXeJteIthewheelbarrowEtEstheoOewheeleIAehEJleoHwhEJhEaMXerHeJtMaster 由helX猜测X为p,有
EMaLOotbeabletoWrowHlowersbPtMLWarIeOproIPJesQPstasMaOLIeaIleaAesolIoAershoespEeJesoHropeaOIbPshelsoHIeaIWrassaOLboILsaOItoIaLEboPWhtawheelbarrowtohelpEOJlearEOWEtPpEhaAealwaLsloAeIaOIrespeJteIthewheelbarrowEtEstheoOewheeleIAehEJleoHwhEJhEaMperHeJtMaster
由EtEn，sbPn知E,P为元音，放入后发现E为i ，P为u,有
iMaLOotbeabletoWrowHlowersbutMLWarIeOproIuJesQustasMaOLIeaIleaAesolIoAershoespieJesoHropeaOIbushelsoHIeaIWrassaOLboILsaOItoIaLibouWhtawheelbarrowtohelpiOJleariOWitupihaAealwaLsloAeIaOIrespeJteIthewheelbarrowitistheoOewheeleIAehiJleoHwhiJhiaMperHeJtMaster
剩下Ｏ和Ｉ最多，猜测为ｄ和ｎ，代人后发现Ｏ为ｎ，Ｉ为ｄ。有
iMaLnotbeabletoWrowHlowersbutMLWardenproduJesQustasManLdeadleaAesoldoAershoespieJesoHropeandbushelsoHdeadWrassanLbodLsandtodaLibouWhtawheelbarrowtohelpinJlearinWitupihaAealwaLsloAedandrespeJtedthewheelbarrowitistheonewheeledAehiJleoHwhiJhiaMperHeJtMaster 由第一句话猜测出M为m,L为y。有
imaynotbeabletoWrowHlowersbutmyWardenproduJesQustasmanydeadleaAesoldoAershoespieJesoHropeandbushelsoHdeadWrassanybodysandtodayibouWhtawheelbarrowtohelpinJlearinWitupihaAealwaysloAedandrespeJtedthewheelbarrowitistheonewheeledAehiJleoHwhiJhiamperHeJtmaster 最后得出明文为 I may not be able to grow flowers .But my garden produces just as many dead leaves, old overshoes pieces of rope and bushels of dead grass anybody sand. Today I bought a wheelbarrow to help in clearing it up. I have alwaysloved and respected the wheelbarrow .It is the one wheel edvehicle of which I am perfect master.
（b）由重合指数法
m=1时 重合指数为 0.043718
m=2时 重合指数为 0.044151，0.052792
m=3时 重合指数为0.064296，0.056601，0.056760
m=4时 重合指数为0.048581，0.054138，0.049036，0.060374
m=5时 重合指数为0.056661，0.057093，0.047004，0.049677，0.057251 m=6时 重合指数为0.079101，0.100128，0.066327，0.081633，0.059949，0.089923 所以密钥长度为6 求Mg比较有
第1位Mg(2)= 0.089923 第2位Mg(17)= 0.070554 第3位Mg(24)= 0.058732 第4位Mg (15) = 0.066000 第5位Mg(19)= 0.055786 第6位Mg(14)=0.070429
密钥K =（2,17,24,15,19,14）
明文为
I learned how to calculate the amount of paper needed for a room when I was at school. you multiply the square foot age of the walls by the cubic contents of the floor and ceiling combined and doubleit you ,then allow half the total for openings such as windows and doors .Then you allow the other half for matching the pattern. Then you double the whole thing again to give a margin of error and then you order the paper. （c）频数为
A = 13 B = 21 C = 32 D = 9 E = 13 F = 10 G = 0 H = 1 I = 16 J = 6 K = 20 L = 0 M = 0 N = 1 O = 2 P = 20 Q = 4 R = 12 S = 1 T = 0 U = 6 V = 4 W = 0 X = 2 Y = 1 Z = 4 所以C解密应为e
设加密为ax+b(mod 26) 4a+b=2(mod 26) 若B解密为t
有19a+b=1(mod 26)
解得a=19,b=4，解密为11y+18(mod 26) 破译后文字为
Ymkxknkdobbonoxycksoehdyxpbyxdocdmosxdnopvoebyxcqvybsoehmkbdyxlbkccksdzybdobvozoosvcksdzybdobvkmbyshdyxrscdysboocdexoozyzoonoczveclbsvvkxdcohzvysdcoddkfkvoebnopysdbowzoozbydoqobkxycpyiobcodxycnbysdc 无意义
若K解密为t
有19a+b=10(mod 26),解得a=4,不合法 若P解密为t
有19a+b=15(mod 26)，解得a=13不合法 若I解密为t
有19a+b=8(mod26),解得a=16不合法 若A解密为t,解得a=12不合法 若E解密为t,解得a=14不合法
若R解密为t,解得a=1,b=24,解密为y+2 破译后文字为
msgtglgderreletmkgcewbdmtxrmtdekdsectdlexhewrmtkqhmrcewbsgrdmtzrgkkgcdfmrderhefeechkgcdfmrderhgsrmcbdmtjckdmcreekdwteefmfeelekfhwkzrchhgtdkebfhmcdkeddgpghewrlexmcdreafeefrmdeqergtmkxmuerkedtmklrmcdk 无意义
若F解密为t，解得a=21,b=22,解密为5x+20(mod 26) 破译后文字为
swobonozerrenebsioueqpzsbvrsbzeizweubznevteqrsbimtsrueqpworzsbfroiiouzjsrzerteje
eutiouzjsrzertowrsupzsbduizsureeizqbeejsjeeneijtqifruttobziepjtsuziezzohoteqrnev
suzrekjeejrszemerobsivsgeriezbsinrsuzi 无意义
若D解密为t，解得a=7,b=0,解密为15x(mod 26) 破译后文字为
ugivifiperrefevuqiaeolpuvdruvpeqpgeavpfedxeoruvqcxuraeolgirpuvhriqqiapturperxete
eaxqiapturperxigrualpuvbaqpuareeqpoveetuteefeqtxoqhraxxivpqeltxuapqeppinixeorfed
uaprewteetrupecerivuqdukerqepvuqfruapq 无意义
a=19，b=4.解密a=11,b=8 附明文
ocanadaterredenosaieuxtonfrontestceintdefleuronsglorieuxcartonbrassaitporterlepe
eilsaitporterlacroixtonhistoireestuneepopeedesplusbrillantsexploitsettavaleurdef
oitrempeeprotegeranosfoyersetnosdroits 加标点后（加拿大国歌法语版） OCanada! Terre de nos a?eux,
Ton front est ceint de fleurons glorieux! Car ton bras sait porter l'épée, Il sait porter la croix! Ton histoire est une épopée Des plus brillants exploits. Et ta valeur de foi trempé
Protégera nos foyers et nos droits; Protégera nos foyers et nos droits (d)
频数统计为
A = 20 B = 25 C = 46 D = 19 E = 17 F = 17 G = 9 H = 7 I = 19 J = 10 K = 30 L = 5 M = 4 N = 5 O = 4 P = 26 Q = 13 R = 22 S = 8 T = 12 U = 9 V = 12 W = 6 X = 6 Y = 5 Z = 7
各个字母均存在而且不少，表明不可能为代换密码和置换密码
根据各种三字母组合，可以判断应该为6字母为一组（相同三字母位置之差大多为6的倍数）但没有找到维吉尼亚密钥。应该是希尔密码（不知道如何破译） 1.22
(a)（刘庆宾） 证明：
方法一(冒泡排序思想类似)：
设Pn~P1的系数依次为Qn’~Q1’，记∑=PnQn’+Pn-1Qn-1’+……+P1Q1’
结论一：如果存在i，j满足i>=j而且Qi’<=Qj’，交换Qi’与Qj’的位置得到 新的∑’，有∑<=∑’。 证明：∑-∑’=
PnQn’+…+PiQi’+…PjQj’+…+P1Q1’-(PnQn’+…+PiQj’+…PjQi’+…+P1Q1’) =(Qi’-Qj’)(Pi-Pj)<=0 假设存在整数k,
(1) k=1时，令Qn=max{ Qn’~Q1’},如果Qn’!=Qn，交换Qn’与Qn的位置，则表达式∑的值增加;否则不交换
(2)k>=2时，我们有Pn~Pk-1的系数依次为Qn~Qk-1. 令Qk=max({Qn’~Q1’}-{ Qn~Qk-1});,同理交换Qk’与Qk的位置，∑的值增加。
由此类推，当k=n时,得到了命题中给出的式子。在证明过程中保证了Qn’~Q1’的任意性，因此当Qn’~Q1’与 Qn~Q1完全对应时，∑达到最大值。
方法二（快速排序思想类似）：
与方法一类似，请感兴趣的朋友自己证明一下
(b)略
1.23(孙宏峰)
解：b r e a t h t a k I n g 1 17 4 0 19 7 19 0 10 8 13 16
R U P O T E N T O I F V 17 20 15 14 19 4 13 19 14 8 5 21 我们可以假设m=2,3,4,……，直到找到密钥为止 若m=2，则有
(1,17)=(17,20) ……………………① (4,0)=(15,14) ……………………② (19,7)=(19,4) (19,0)=(13,19) (10,8)=(14,8) (13,6)=(5,21)
由①②可得K=，可以解出K，然后代入其他式中验算K的正确性，同理，可以
求得若m=3，4,……时的密钥K并进行验算，最终求出正确的密钥K