【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)
(2015-04-28, WORD, 3KB, 下载1次)
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