文件名称:CBVBBBVJRNRGRJ
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题目:一个5位数,判断它是不是回文数。即12321是回文数,个位与万位相同,十位与千位相同。
1.程序分析
2.程序源
代码如下:
#include stdio.h
#include conio.h
main( )
{
long ge,shi,qian,wan,x
scanf( ld ,&x)
wan=x/10000
qian=x 10000/1000
shi=x 100/10
ge=x 10
if(ge==wan&&shi==qian) 个位等于万位并且十位等于千位
printf( this number is a huiwen\n )
else
printf( this number is not a huiwen\n )
getch()
}-Title: a 5-digit, determine if it is a palindrome. That 12321 is a palindrome, the same bits and ten thousand, ten and one thousand identical.
1. Program analysis
2. Source
Code is as follows:
#include stdio.h
#include conio.h
main ()
{
long ge, shi, qian, wan, x
scanf ( ld , & x)
wan = x/10000
qian = x 10000/1000
shi = x 100/10
ge = x 10
if (ge == wan && shi == qian) a position equivalent to ten thousand and ten equal to one thousand* /
printf ( this number is a huiwen \ n )
else
printf ( this number is not a huiwen \ n )
getch ()
}
1.程序分析
2.程序源
代码如下:
#include stdio.h
#include conio.h
main( )
{
long ge,shi,qian,wan,x
scanf( ld ,&x)
wan=x/10000
qian=x 10000/1000
shi=x 100/10
ge=x 10
if(ge==wan&&shi==qian) 个位等于万位并且十位等于千位
printf( this number is a huiwen\n )
else
printf( this number is not a huiwen\n )
getch()
}-Title: a 5-digit, determine if it is a palindrome. That 12321 is a palindrome, the same bits and ten thousand, ten and one thousand identical.
1. Program analysis
2. Source
Code is as follows:
#include stdio.h
#include conio.h
main ()
{
long ge, shi, qian, wan, x
scanf ( ld , & x)
wan = x/10000
qian = x 10000/1000
shi = x 100/10
ge = x 10
if (ge == wan && shi == qian) a position equivalent to ten thousand and ten equal to one thousand* /
printf ( this number is a huiwen \ n )
else
printf ( this number is not a huiwen \ n )
getch ()
}
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