说明：  【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)
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