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Description: 统计输入文件的行数、空格数、元音字母数、辅音字母数、非空白字母数、非空的行数-statistical inputs the line a few, a few spaces, a few vowel and consonant few letters, a few non-blank letters, non-empty rows
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Size: 1277 |
Author: 中文名 |
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Description: 语音信号处理中的端点检测，用于清浊音的鉴别。是语音预处理中的必要过程-The detection of end point in speech signal processing is utilized for distinguishing vowel and sonant
Platform: |
Size: 249402 |
Author: 陈为 |
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Description: 统计输入文件的行数、空格数、元音字母数、辅音字母数、非空白字母数、非空的行数-statistical inputs the line a few, a few spaces, a few vowel and consonant few letters, a few non-blank letters, non-empty rows
Platform: |
Size: 1024 |
Author: 中文名 |
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Description: 语音信号处理中的端点检测，用于清浊音的鉴别。是语音预处理中的必要过程-The detection of end point in speech signal processing is utilized for distinguishing vowel and sonant
Platform: |
Size: 248832 |
Author: 陈为 |
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Description: 实现一个简单的密码生成系统。一个有效的密码由L(3<=L<=15)个小写字母(来自传统的拉丁字母集 a ... z )组成，至少有一个元音字母( a , e , i , o 或者 u )，至少有两个辅音字母(除去元音以外的字母)，并且按字母表顺序出现(例如， acd 是有效的，而 dac 不是)。
给定一个期望长度L和若干个小写字母，写一个程序，打印出所有的长度为L、能由这些字母组成的有效密码。密码必须按字母表顺序打印出来-The realization of a simple code generation system. A valid password from the L (3 < = L < = 15) a lowercase letters (from the Latin tradition set ' a' ...' z' ) composed of at least one vowel letter ( ' a' , ' e' , ' i' , ' o' or ' u' ), at least two consonant letters (letters other than vowels removed), and appear in alphabetical order (for example, ' acd' is valid, and ' dac ' instead). Expectations of a given length L and a number of lowercase letters, writing a program, print out all of the length L, can be composed by these letters effectively password. Password must be printed in alphabetical order
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Size: 1024 |
Author: steven |
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Description: 1.Takes a number and return the factorial of it.
2. Decimal input output.
3. Vowel and consonant count.
All are Implemented in Assembly.
Platform: |
Size: 2048 |
Author: shi |
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Description: "This game will get character and do in these steps.\n"+
"1)Get only character from a-z (no capital letters)\n"+
"2)After that, it will reverse the result from step 1)\n"+
"3)Finally, it will find vowel from step 1) (before reverse) and append after result from step 2) (after reverse)" -"This game will get character and do in these steps.\n"+
"1)Get only character from a-z (no capital letters)\n"+
"2)After that, it will reverse the result from step 1)\n"+
"3)Finally, it will find vowel from step 1) (before reverse) and append after result from step 2) (after reverse)"
Platform: |
Size: 1024 |
Author: Netz |
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Description: 对于元音字母（a e i o u）开头的英文单词结尾加上cow；相反，如果是辅音开头的英文单词，开头第一个字母移到末尾加ow-Bovine Latin
The cows have heard that the pigs use a secret language called "Pig Latin" when they want to communicate with each other without Farmer John being able to understand what they are saying. Thinking this is an excellent idea, they have invented their own version, aptly called Bovine Latin.
Converting an English word to a Bovine Latin word is quite simple. For words that start with a vowel ( a , e , i , o or u ), "cow" is added to the end of the word for example, "udder" becomes "uddercow". For words that do not begin with a vowel, the first letter is moved to the end of the word, and "ow" is added e.g., "farmer" becomes "armerfow". So "the cows escape at dawn" becomes "hetow owscow escapecow atcow awndow." The cows fervently believe that FJ will not understand this subterfuge.
Never known as enthusiastic linguists, the cows find this translation quite tedious and thus have asked you to write a program that will take single words and translate them into Bovine L
Platform: |
Size: 1024 |
Author: adonis |
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Description: 实现一个简单加解密系统，要求如下：
(1)输入要加密的原文，长度不小于30个字母；
(2)输入密钥长度L及用来产生密钥的英文字母(个数>=L)；
注：一个有效的密钥由L(3<=L<=15)个小写英文字母组成，至少有一个元音字母( a , e , i , o 或者 u )，至少有两个辅音字母(除去元音以外的字母)。
(3)选择加密功能，可对原文进行加密；
(4)选择解密功能，可对密文进行解密。 -Implement a simple encryption and decryption system requirements are as follows: (1) Enter the encrypted text, the length is not less than 30 letters (2) Enter the key length L and used to generate the key in English letters (number> = L ) Note: a valid key from the L (3 < = L < = 15) composed of a lower-case letters, at least one vowel letter ( ' a' , ' e' , ' i' , ' o' or ' u' ), at least two consonant letters (other than the removal of vowel letters). (3) Select encryption can encrypt the original text (4) Select decryption functions, can decrypt the ciphertext.
Platform: |
Size: 1024 |
Author: gqw |
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Description: Vowel extraction using HU moments of binary image in matlab
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Size: 8192 |
Author: ddokok |
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Description: 输入一句英语句子，删除单词里连续重复的单词和元音字母。如果单词只有一个字母不删除。-Enter an English sentence, delete the word where a continuous repetition of words and vowel letters. If the word is only one letter is not removed.
Platform: |
Size: 1024 |
Author: 张潮铭 |
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Description: this a vowel program using java -this is a vowel program using java
Platform: |
Size: 1024 |
Author: mpk |
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Description: A cross-language vowel normalisation procedure using matlab.
Platform: |
Size: 3072 |
Author: STARRY |
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Description: lex program for vowel constantn
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Size: 9216 |
Author: shashi |
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Description: We propose methods to track natural variations
in the characteristics of the vocal-tract system from speech
signals. We are especially interested in the cases where these
characteristics vary over time, as happens in dynamic sounds
such as consonant-vowel transitions. We show that the selection
of appropriate analysis segments is crucial in these methods, and
we propose a selection based on estimated instants of significant
excitation. These instants are obtained by a method based on
the average group-delay property of minimum-phase signals.
In voiced speech, they correspond to the instants of glottal
closure. The vocal-tract system is characterized by its formant
parameters, which are extracted from the analysis segments.
Because the segments are always at the same relative position
in each pitch period, in voiced speech the extracted formants are
consistent across successive pitch periods. We demonstrate the
results of the analysis for several difficult cases of speech signals.
Platform: |
Size: 263168 |
Author: emerix |
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Description: 关于语音识别方面的论文，基于HMM模型，有利于初学者掌握-A thesis on speech recognition, HMM-based model is beneficial for beginners to master
Platform: |
Size: 217088 |
Author: 吴威麒 |
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Description: 问题的提出：17世纪法国数学家加斯帕的一本《数学的游戏问题》描述了许多有趣问题，约瑟问题就是其中之一：
15名基督教徒和15名异教徒同乘一船航行，途中风浪大作，危机万分，领航者告诉大家，只要将全船的一半人投入海中，其余人就能幸免。大家都同意这个办法，并协定者30人围成一圈；由第一个人起报数，每数至第9人便把他投入海中，下一个接着从1开始报数，第9人又被投入海中，依次循环，直至剩下15人为止。问题是如何排法使投入海中的人全为异教徒？
编程的思路：此问题有人把它的答案隐示于下列詩句中：
From numbers’ aid and art, (依靠数字的和魔力，)
Never will fame depart! （声誉决不会离你而去！）
这句诗中元音字母依次为：oueaiaaeeiaeea，分别用1、2、3、4、5代替a、e、i、o、u，便得一排数（划线得数字表示基督教徒，不划线的表示异教徒）。
4 5 2 1 3 1 1 2 2 3 1 2 2 1 故所求排法诗4个基督教徒，5个异教徒，再2个基督教徒，1个异教徒，…可以证明，这种排法所有的基督教徒都是安全的。
-Of the problem: the 17th century French mathematician Gaspard s a "mathematical game problem " describes the many interesting questions, Joseph is one of them:
15 Christians and 15 pagans sail with the boat ride, the way storm comes up, extreme crisis, the leader tell you that as long as half of the whole ship into the sea, the rest can be spared. We all agree with this approach, and there were 30 agreements circle the first person from the reported number, the number to 9 for each person put into the sea in his next start and then reported from a number of people has been put into 9 The sea, in cycles, until the remaining 15 people so far. The question is how to arrange method to make the people into the sea are all infidels?
Programming idea: the problem was the answer to it implicitly in the following verse:
From numbers aid and art, (relying on numbers and magic)
Never will fame depart! (reputation will never leave you!)
Poem vowel sentence were: oueaiaaeeiaeea, were re
Platform: |
Size: 12288 |
Author: yinlingzhi |
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Description: speech data set.vowel regonition
Platform: |
Size: 137216 |
Author: narges |
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Description: vowel _phenome _classification
Platform: |
Size: 3822592 |
Author: bhai |
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Description: this function is used to get formant frequencies and vocal models for one vowel and one speaker
code is correctly commented to take in one trial, can be expanded to five by uncommenting-this function is used to get formant frequencies and vocal models for one vowel and one speaker
code is correctly commented to take in one trial, can be expanded to five by uncommenting
Platform: |
Size: 1024 |
Author: kimmy |
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