c - 逆序/正序输出每位.

 1 #include <stdio.h>

 2 #include <math.h>

 3 

 4 /*

 5 判断一个正整数的位数,并按正序,逆序输出他们的位.

 6 */

 7 

 8 int 

 9 invert(int);

10 

11 void

12 order(int, int);

13 

14 int 

15 main(void) {

16     int n = 56789;

17     printf("original:%d\n", n);

18     int bitCont = invert(n);

19     printf("\nbits: %d\n", bitCont);

20     order(n, bitCont);

21     return 0;

22 }

23 

24 //逆序输出每一位并记录位数.

25 int 

26 invert(int n) {

27     int bitCount = 0;    //记录位数.

28     do 

29     {

30         ++bitCount;

31         printf("%d ", n % 10);    //逆序输出每一位.

32         n /= 10;

33     } while (n);    //终止条件是"n为0".

34     return bitCount;

35 }

36 

37 //正序输出每一位.

38 void

39 order(int n, int bitCount) {

40     int tmp = 0;    //存储即将减去的级别,如当n为"1234"时,存储"1000";当n为"234"时,存储"100".

41     int h = 0;    //存储最高位的位.

42     while(n) {

43         tmp = (int)pow((double)10, --bitCount);

44         h = n / tmp;

45         printf("%d ", h);

46         n -= h * tmp;

47     }

48 }

 

output:

original:56789

9 8 7 6 5

bits: 5

5 6 7 8 9 请按任意键继续. . .

 

 1 //二,使用(求余数方法)

 2 void 

 3 count_vMod(int input) {

 4     //分别表示'个位,十位,百位,千位,万位'上的数字.

 5     int one, ten, hundred, thousand, ten_thousand;

 6     

 7     ten_thousand = input / 10000;    //除1万.

 8     thousand = input % 10000 / 1000;    //取余1万后除1千.

 9     hundred = input % 1000 / 100;    //取余1千后除1百.

10     ten = input % 100 / 10;    //取余1百后除1十.

11     one = input % 10;

12 

13     if(ten_thousand)

14         printf("共有5位数,%d %d %d %d %d\n", one, ten, hundred, thousand, ten_thousand);

15     else if(!ten_thousand && thousand)

16         printf("共有4位数,%d %d %d %d\n", one, ten, hundred, thousand);

17     else if(!ten_thousand && !thousand && hundred)

18         printf("共有3位数,%d %d %d\n", one, ten, hundred);

19     else if(!ten_thousand && !thousand && !hundred && ten)

20         printf("共有2位数,%d %d\n", one, ten);

21     else if(!ten_thousand && !thousand && !hundred && !ten && one)

22         printf("共有1位数,%d\n", one);

23 }

虽然第二中方法(求余数)在求某位数的时候,有用,但是感觉比较笨拙(主要是对这个题目,特别是if判断时).