3. 求循环小数

分手后的思念是犯贱 2022-04-11 09:58 141阅读 0赞

对于任意的真分数 N/M （ 0 < N < M ），均可以求出对应的小数。如果采用链表表示各个小数，对于循环节采用循环链表表示，则所有分数均可以表示为如下链表形式。

![循环节][E5_BE_AA_E7_8E_AF_E8_8A_82.JPG]

**输入：** N M

**输出：** 转换后的小数（不超过 50 ）

**要求：** 仅编写将分数转换为小数的函数 change( int n, int m, NODE \* head ) 。

### 预设代码 ###

前置代码

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1.  /\* PRESET CODE BEGIN - NEVER TOUCH CODE BELOW \*/
2.  
3.  \#include <stdio.h>
4.  \#include <stdlib.h>
5.  
6.  typedef struct node
7.  \{ int data;
8.  struct node \* next;
9.  \} NODE;
10. 
11. void output( NODE \* );
12. void change( int, int, NODE \* );
13. 
14. void output( NODE \* head )
15. \{ int k=0;
16. 
17. printf("0.");
18. while ( head->next != NULL && k<50 )
19. \{ printf("%d", head->next->data );
20. head = head->next;
21. k ++;
22. \}
23. printf("\\n");
24. \}
25. 
26. int main()
27. \{ int n, m;
28. NODE \* head;
29. 
30. scanf("%d%d", &n, &m);
31. head = (NODE \*)malloc( sizeof(NODE) );
32. head->next = NULL;
33. head->data = -1;
34. change( n, m, head );
35. output( head );
36. return 0;
37. \}
38. 
39. /\* PRESET CODE END - NEVER TOUCH CODE ABOVE \*/

#include <stdio.h>  
    #include <stdlib.h>  
      
    typedef struct node  
    {   int         data;  
        struct node * next;  
    } NODE;  
      
    void output( NODE * );  
    void change( int, int, NODE * );  
      
    void output( NODE * head )  
    {   int k=0;  
      
        printf("0.");  
        while ( head->next != NULL && k<50 )  
        {   printf("%d", head->next->data );  
            head = head->next;  
            k ++;  
        }  
        printf("\n");  
    }  
      
    int main()  
    {   int n, m;  
        NODE * head;  
      
        scanf("%d%d", &n, &m);  
        head = (NODE *)malloc( sizeof(NODE) );  
        head->next = NULL;  
        head->data = -1;  
        change( n, m, head );  
        output( head );  
        return 0;  
    } 
    
    void change( int n, int m, NODE * head ){
        NODE *p, *q;
        int i = 0, x = 0, yushu = 1;
        n = n * 10;
        x = n / m;
        p = head;
        q = (NODE *)malloc(sizeof(NODE));
        q->data = x;
        p->next = q;
        p = q;
        while(yushu && i<50){
            n = n - x * m;
            n = n * 10;
            x = n / m;
            q = (NODE *)malloc(sizeof(NODE));
            q->data = x;
            p->next = q;
            p = q;
            i++;
            yushu = n % m;
        }
        p->next = NULL;
    }

[E5_BE_AA_E7_8E_AF_E8_8A_82.JPG]: http://online.bit.edu.cn/moodle/pluginfile.php/185551/mod_programming/intro/%E5%BE%AA%E7%8E%AF%E8%8A%82.JPG
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