【程序29】

题目：给一个不多于5位的正整数，要求：一、求它是几位数，二、逆序打印出各位数字。

1. 程序分析：学会分解出每一位数，如下解释：(这里是一种简单的算法，师专数002班赵鑫提供)

2.程序源代码：

复制代码 代码如下:

#include stdio.h

#include conio.h

main( )

{

long a,b,c,d,e,x

scanf( ld ,&x)

a=x/10000 分解出万位

b=x 10000/1000 分解出千位

c=x 1000/100 分解出百位

d=x 100/10 分解出十位

e=x 10 分解出个位

if (a!=0) printf( there are 5, ld ld ld ld ld\n ,e,d,c,b,a)

else if (b!=0) printf( there are 4, ld ld ld ld\n ,e,d,c,b)

else if (c!=0) printf( there are 3, ld ld ld\n ,e,d,c)

else if (d!=0) printf( there are 2, ld ld\n ,e,d)

else if (e!=0) printf( there are 1, ld\n ,e)

getch()

}

-Topic: no more than five to one positive integer requirements: First, find it is several orders, two, reverse print out the digits.

1. Program analysis: the decomposition of each digit Society

2. Source Code:

Code is as follows:

#include　stdio.h

#include　conio.h

main ()

{

   　long a, b, c, d, e, x

   　scanf (　 ld , & x)

   　a = x/10000　　break out ten thousand* /

   　b = x　10000/1000　　break out one thousand* /

   　c = x　1000/100　　break out one hundred* /

   　d = x　100/10　　break out ten* /

   　e = x　10　　break out bits* /

   　if (! a = 0) printf ( there are 5,　ld　ld　ld　ld　ld \ n , e, d, c, b, a)

   　else if (! b = 0) printf ( there are 4,　ld　ld　ld　ld \ n , e, d, c, b)

       　else if (! c = 0) printf ( there are 3,　ld　ld　ld \ n , e, d, c)

           　else if (! d = 0) printf ( there are 2,　ld　ld \ n , e, d)