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find_huiwen
- 随机输入一个数,判断它是不是对称数(回文数)(如3,121,12321,45254)。不能用字符串库函数-Enter a random number to determine the number it is not symmetrical (palindromic) (as 3,121,12321,45254).Can not use the string library functions
Code5
- 判断一个数是否为回文数,提供了两种方法以便参考,方法一:只要利用内置函数解决该问题,方法二:主要通过数组判断-Determine if a number is palindrome, offers two ways to reference one: as long as the use of built-in functions to solve the problem, method two: mainly through the array to determine
Number
- 输入一个1至99999之间的数,再求在1至99999之间的回文数-Enter a number between 1-99999, and then seek from 1 to 99999 of palindromic
Palindrome-number
- 编写一个java应用程序。用户从键盘输入一个1-9999之间的数,程序将判断这个数是几位数,并判断这个数是否回文数。回文数是指将数含有的数字逆序排列后得到的数和原数相同,例如12121,4224,6778776等都是回文数。 1)程序具有判断用户的输入是否为合法整数的功能。对非法输入(例如含有字母)要进行处理。 2)要判断输入数的位数,并输出相关信息。 3)要判断是否回文数。 -Write a Java applications. The user from the keybo
dfdfdsd
- 判断回文数,很不错的易语言源码,易爱好者可以下载使用-Determine palindrome, a very good source of easy language, easy to use fans can download
huiwenshu
- 回文数 回文数 回文数 回文数 回文数 回文数-Palindromic function Palindromic functionPalindromic functionPalindromic functionPalindromic functionPalindromic functionPalindromic function
HomeWork1
- 编写一个java应用程序。用户从键盘输入一个1-9999之间的数,程序将判断这个数是几位数,并判断这个数是否回文数。回文数是指将数含有的数字逆序排列后得到的数和原数相同,例如12121,4224,6778776等都是回文数。-Write a java application. User from the keyboard to enter a number between 1-9999, the program will determine this number is several numb
PalindromeTester
- 回文数,忽略标点和空格,测试回文数的好代码-Palindrome, ignoring punctuation and spaces, a good number of test code for palindrome
huiwen
- 回文序列 编写一个程序 来判断所输入的数字是否为回文数 回文数是指如:98789, 这个数字正读是98789,倒读也是98789,正读倒读一样,所以这个数字就是回文数。-hui wen shu xu lie
cs98876
- 易语言源码分享。希望更多的朋友了解易语言。源码名称:判断回文数。体验中文编程-Easy language source code sharing.Hope more friends to understand easy language.Source Name: judgment palindrome number.Chinese programming experience
huiwen
- 判断从键盘输入的一个数是否是回文数的小程序。-palindrome data
huiwenshu
- C语言 回文数经典算法 巧用while语句-Palindrome
huiwen
- 对回文数进行管理,abccba以及abcba都是回文数。-Palindrome the management, abccba, as well as abcba palindrome.-
chengxu
- 求11--999之间的回文数 比较实用 大家可以放心下载-failed to translate
huiwenshu
- 用Vc++编写的回文数,,一个简单的VC++源代码,希望大家用得上-With Vc++ to write the number of palindromes, a simple VC++ source code, we need them
c
- C语言的一些简单经常用到的小工具程序段,比如说冒泡排序,回文数,素数分解等等。-Some of the C language simple and often use the gadget program segments, such as bubble sort, palindrome number, prime number decomposition and so on.
a
- 一个数的各位数字倒过来所得到的新数叫原数的反序数. 如果一个数等于它的反序数,则称它为对称数或回文数。 求10000以内的二进制对称数.-The digits of a number of upside down to get a new number called the anti-ordinal number of the original number. If a number is equal to its inverse ordinal number, call it th
Looking-for-Palindrome-made-several
- 寻找并输出11~999之间的数m,它满足m、m的平方、m的立方均为回文数。所谓回文数是指其各位数字左右对称的整数,例如121、676、94249等。满足上述条件的数如 m=11,其平方为121,其立方为1331,皆为回文数。要求编制函数 int svalue(long m)实现此功能,如果是回文数,则函数返回1,反之则返回0。在主函数中将三重回文数输出。-Looking for and output 11 ~ 999 between several m, it satisfies m, m sq
huiwen
- 测试从键盘上输入的一个规定范围内的数判断是否为回文数-The number of test inputs to whether the number of reported
palindromic-number
- 如果一个数从左边和从右边读都是相同的数,就称它为回文数,例如383,求出500以内的回文数并输出显示。 -If a number from the left and from the right to read is the same number, he called it reported number, such as 383, and from 500 the number of reported within and the output shown.