1个回答

Substitution Cipher 程序的设计
Description Antique Comedians of Malidinesia would like to play a new discovered comedy of Aristofanes. Putting it on a stage should be a big surprise for the audience so all the preparations must be kept absolutely secret. The ACM director suspects one of his competitors of reading his correspondece. To prevent other companies from revealing his secret, he decided to use a substitution cipher in all the letters mentioning the new play. Substitution cipher is defined by a substitution table assigning each character of the substitution alphabet another character of the same alphabet. The assignment is a bijection (to each character exactly one character is assigned -- not neccessary different). The director is afraid of disclosing the substitution table and therefore he changes it frequently. After each change he chooses a few words from a dictionary by random, encrypts them and sends them together with an encrypted message. The plain (i.e. non-encrypted) words are sent by a secure channel, not by mail. The recipient of the message can then compare plain and encrypted words and create a new substitution table. Unfortunately, one of the ACM cipher specialists have found that this system is sometimes insecure. Some messages can be decrypted by the rival company even without knowing the plain words. The reason is that when the director chooses the words from the dictionary and encrypts them, he never changes their order (the words in the dictionary are lexicographically sorted). String a1a2 ... ap is lexicografically smaller than b1b2 ... bq if there exists an integer i, i <= p, i <= q, such that aj=bj for each j, 1 <= j < i and ai < bi. The director is interested in which of his messages could be read by the rival company. You are to write a program to determine that. Input The input consists of N cases. The first line of the input contains only positive integer N. Then follow the cases. The first line of each case contains only two positive integers A, 1 <= A <= 26, and K, separated by space. A determines the size of the substitution alphabet (the substitution alphabet consists of the first A lowercase letters of the english alphabet (a--z) and K is the number of encrypted words. The plain words contain only the letters of the substitution alphabet. The plain message can contain any symbol, but only the letters of the substitution alphabet are encrypted. Then follow K lines, each containing exactly one encrypted word. At the next line is encrypted message. Output For each case, print exactly one line. If it is possible to decrypt the message uniquely, print the decrypted message. Otherwise, print the sentence 'Message cannot be decrypted.'. Sample Input 2 5 6 cebdbac cac ecd dca aba bac cedab 4 4 cca cad aac bca bdac Sample Output abcde Message cannot be decrypted.
"0x53d21cca"指令引用的"0x00000005"内存
Visual Basic6.0使用DockStudioXP 2.0控件以后关闭窗口出现"0x53d21cca"指令引用的"0x00000005"内存

SVN迁移到GIT报Checksum mismatch: TapiParser/jar/tapiParser/tapiParser.jar c7a0bbbcc00e06bf963e4499faf8292c29133090 expected: c8d8b55b252bad17d6e53a6435eb44cd got: 706a1cb8c2fe7c059d37cbf427df4795

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"http-bio-8080-exec-192" daemon prio=10 tid=0x00007f8468047000 nid=0x4c47 waiting for monitor entry [0x00007f84f8cca000] java.lang.Thread.State: BLOCKED (on object monitor) at java.lang.Class.forName0(Native Method) at java.lang.Class.forName(Class.java:274) at com.sun.beans.finder.ClassFinder.findClass(ClassFinder.java:67) at com.sun.beans.finder.ClassFinder.findClass(ClassFinder.java:110) at java.beans.Introspector.findCustomizerClass(Introspector.java:1245) at java.beans.Introspector.getTargetBeanDescriptor(Introspector.java:1239) at java.beans.Introspector.getBeanInfo(Introspector.java:415) at java.beans.Introspector.getBeanInfo(Introspector.java:252) at java.beans.Introspector.getBeanInfo(Introspector.java:214) at com.java.manage.util.JsonUtils.bean2json(JsonUtils.java:76) at com.java.manage.util.JsonUtils.object2json(JsonUtils.java:55) at com.java.manage.util.JsonUtils.list2json(JsonUtils.java:133) at com.java.manage.util.JsonUtils.object2json(JsonUtils.java:49) at com.java.manage.util.JsonUtils.bean2json(JsonUtils.java:84) at com.java.manage.webservice.GroupMemberManager.getContactsList(GroupMemberManager.java:409) at com.java.manage.webservice.GetEnterMouthWebService.get(GetEnterMouthWebService.java:111) at sun.reflect.GeneratedMethodAccessor74.invoke(Unknown Source) at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43) at java.lang.reflect.Method.invoke(Method.java:606) at com.sun.jersey.spi.container.JavaMethodInvokerFactory\$1.invoke(JavaMethodInvokerFactory.java:60) at com.sun.jersey.server.impl.model.method.dispatch.AbstractResourceMethodDispatchProvider\$TypeOutInvoker._dispatch(AbstractResourceMethodDispatchProvider.java:185) at com.sun.jersey.server.impl.model.method.dispatch.ResourceJavaMethodDispatcher.dispatch(ResourceJavaMethodDispatcher.java:75)
C语言算法，Substitution Cipher怎么做
Description Antique Comedians of Malidinesia would like to play a new discovered comedy of Aristofanes. Putting it on a stage should be a big surprise for the audience so all the preparations must be kept absolutely secret. The ACM director suspects one of his competitors of reading his correspondece. To prevent other companies from revealing his secret, he decided to use a substitution cipher in all the letters mentioning the new play. Substitution cipher is defined by a substitution table assigning each character of the substitution alphabet another character of the same alphabet. The assignment is a bijection (to each character exactly one character is assigned -- not neccessary different). The director is afraid of disclosing the substitution table and therefore he changes it frequently. After each change he chooses a few words from a dictionary by random, encrypts them and sends them together with an encrypted message. The plain (i.e. non-encrypted) words are sent by a secure channel, not by mail. The recipient of the message can then compare plain and encrypted words and create a new substitution table. Unfortunately, one of the ACM cipher specialists have found that this system is sometimes insecure. Some messages can be decrypted by the rival company even without knowing the plain words. The reason is that when the director chooses the words from the dictionary and encrypts them, he never changes their order (the words in the dictionary are lexicographically sorted). String a1a2 ... ap is lexicografically smaller than b1b2 ... bq if there exists an integer i, i <= p, i <= q, such that aj=bj for each j, 1 <= j < i and ai < bi. The director is interested in which of his messages could be read by the rival company. You are to write a program to determine that. Input The input consists of N cases. The first line of the input contains only positive integer N. Then follow the cases. The first line of each case contains only two positive integers A, 1 <= A <= 26, and K, separated by space. A determines the size of the substitution alphabet (the substitution alphabet consists of the first A lowercase letters of the english alphabet (a--z) and K is the number of encrypted words. The plain words contain only the letters of the substitution alphabet. The plain message can contain any symbol, but only the letters of the substitution alphabet are encrypted. Then follow K lines, each containing exactly one encrypted word. At the next line is encrypted message. Output For each case, print exactly one line. If it is possible to decrypt the message uniquely, print the decrypted message. Otherwise, print the sentence 'Message cannot be decrypted.'. Sample Input 2 5 6 cebdbac cac ecd dca aba bac cedab 4 4 cca cad aac bca bdac Sample Output abcde Message cannot be decrypted.
windows server 2003系统组件运行报错？

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2015-02-03 02:40:49.816 Gifted[7798:60b] You've implemented -[<UIApplicationDelegate> application:didReceiveRemoteNotification:fetchCompletionHandler:], but you still need to add "remote-notification" to the list of your supported UIBackgroundModes in your Info.plist. 2015-02-03 02:40:49.828 Gifted[7798:60b] Multi-tasking -> Device: YES, App: YES 2015-02-03 02:40:50.086 Gifted[7798:60b] *** Terminating app due to uncaught exception 'CALayerInvalidGeometry', reason: 'CALayer position contains NaN: [nan nan]' *** First throw call stack: (0x2fed1ecb 0x3a668ce7 0x2fed1e0d 0x323773bb 0x323772bf 0x3237724f 0x32700443 0x32714563 0x3271449f 0x1661b3 0x165c81 0x1653f5 0x1664d3 0x164b65 0x1716cf 0x171b31 0x16e83b 0xf6757 0x32707a33 0x327077f1 0x3270e469 0x3270bdb9 0x32775a31 0xf529b 0x32772ca1 0x327725f3 0x3276cca9 0x32708c77 0x32707dd9 0x3276c3e5 0x34d65b55 0x34d6573f 0x2fe9c807 0x2fe9c7a3 0x2fe9af6f 0x2fe05729 0x2fe0550b 0x3276b60b 0x32766871 0xf34b3 0xf3440) libc++abi.dylib: terminating with uncaught exception of type NSException (lldb)
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0 CoreFoundation 0x0000000182d16e50 <redacted> + 148, 1 libobjc.A.dylib 0x000000018237bf80 objc_exception_throw + 56, 2 CoreFoundation 0x0000000182d16d08 <redacted> + 0, 3 Foundation 0x000000018369c124 <redacted> + 112, 4 UIKit 0x000000018810aad0 <redacted> + 3172, 5 UIKit 0x0000000188107674 <redacted> + 168, 6 FrontBoardServices 0x00000001846b77ac <redacted> + 36, 7 FrontBoardServices 0x00000001846b7618 <redacted> + 168, 8 FrontBoardServices 0x00000001846b79c8 <redacted> + 56, 9 CoreFoundation 0x0000000182ccd124 <redacted> + 24, 10 CoreFoundation 0x0000000182cccbb8 <redacted> + 540, 11 CoreFoundation 0x0000000182cca8b8 <redacted> + 724, 12 CoreFoundation 0x0000000182bf4d10 CFRunLoopRunSpecific + 384, 13 UIKit 0x0000000187ecf834 <redacted> + 460, 14 UIKit 0x0000000187ec9f70 UIApplicationMain + 204, 15 XM030 0x00000001003ece1c main + 176, 16 libdyld.dylib 0x00000001827928b8 <redacted> + 4 ) , Application windows are expected to have a root view controller at the end of application launch, NSInternalInconsistencyException ) 正常登陆不崩溃
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java源码分析 Arrays.asList()与Collections.unmodifiableList()
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