第二次期中考補考-第二題

3 like 0 dislike
2.7k views

若一個正整數的奇數位數字和與偶數位數字和兩者差的絕對值是11的倍數，則這個正整數就是11的倍數，否則就不是11的倍數。

請寫一個程式來判斷輸入的數是不是十一的倍數。

 

輸入說明：

會輸入一個正整數N，N不會超過1000位。

 

輸出說明：

如果輸入的數字是11的倍數就在螢幕上輸出YES，否則就輸出NO。

 

輸入範例：
 

11111111111111111111111111111111111111111111111111



 

輸出範例：

 

YES
[Exam] asked in 2017-1 程式設計(一)AD by (30k points)
ID: 38939 - Available when: 2017-12-23 09:00 - Due to: 2017-12-23 12:10 | 2.7k views

16 Answers

0 like 0 dislike 
Hidden content!
#include<stdio.h>

int main()
{
   **    *   ** *      *** **** arr[1000];
* *  *     ***  * *  **  ** * even=0,odd=0,i,n,sum;
    
  ** *** * **   ** *  ** * * ** *** *   *    ** *    *  **
    
* *** ** ***   **     **  *   
   *    * **  ***** * *   * 
  **  ***   **  *     *     * **   **   ***** * 
    {
 **  * *    *    *** ****  *   * ****   * * *    *   *   * * **  *    *    **  * *   **  *     ** * ** ** *   *  ** *   * **** *   * **   ***   
* * **    ** **   *  * ** ********* * * ** **    **** *** * *     *  ** * *  * *   ****   * **   **  ***      *     * ***   *   *     
 *  * ***   *    ** *  *  **  **  *    * *   ****      ** ******   ***       ** *** *    * * * * *** *   *    **** * *  ** ** *  **   * *  *** *
*   * * * *  * * **    ** * *  *  ****** ** * ** **   *****     ***  * *  *  *  * *  * ***** ***** ******  **   *  **    ** * ** *** ***  **
    }
    
***  * ** * **** *  *** * ** *  
    *** ** * * **** *** **    
  ****    * *   **   *  * * ** *  ***     *    *
****     * * **** *** *** **  ***  * 
* **  *  *****  *  * ** *   * ***       ** ** **** ** *  
    
  * ** ** *  * *  * * **** *  0;
}
answered by (215 points)
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
Case 3: Wrong output
Case 4: Wrong output
Case 5: Wrong output
Case 6: Wrong output
Case 7: Wrong output
commented by (4.3k points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
Case 3: Correct output
Case 4: Correct output
Case 5: Correct output
Case 6: Correct output
Case 7: Correct output
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
#include<stdio.h>
int main()
{
*** *****    ** **   *  ** num[1000]={0};
    int k, even, odd, i, j;
** **  *     ** **** *  *** ******    **   *     **   *
    **** ***   *  * ** *    *  *  
 *   *    * **     * * *  *** *
 ** ***** *   *  * * ***  * *   *  *  ***** ****  *    *  **  **  
 ****   * **   * **  *  ****     **  **  ** ***** *
* * * * ** * *       *    ** *  **  *** *  ** * * **** 
* **  *   **   ***  *  * * *   *  *       ***
****   * *****   ** *** ** ***** *  ***
*   * **** ****** **  *      * * ** ** ** ** ** *** ****
** *** ********* *  * * * ***      
 *  * * ****  *    **** ** *        ** * **  * *   
* **  *    **  * ***    * *  **  0;
}
answered by (144 points)
edited by
0 0
prog.c: In function 'main':
prog.c:5:23: error: expected expression before ']' token
     scanf("%d",&array[]);
                       ^
prog.c:7:5: error: 'returm' undeclared (first use in this function)
     returm 0;
     ^~~~~~
prog.c:7:5: note: each undeclared identifier is reported only once for each function it appears in
prog.c:7:12: error: expected ';' before numeric constant
     returm 0;
            ^
commented by (4.3k points)
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
Case 3: Wrong output
Case 4: Wrong output
Case 5: Wrong output
Case 6: Wrong output
Case 7: Wrong output
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
#include<stdio.h>
#include<string.h>
int main()
{
    char c[1000];
 *** *** *  **********   **     *** ***  *** *  
    int a,i,x=0,y=0,z;
*** *    ******   ** ***    **
**  ** ** *** *  **    *****  *        * ** *
    {
*          *  **  ***     * ** ***  * *      ***  *  
   * *  * ** **       **  ***      *  * *    ** *     *  *
  * *     ***    *    * **** ***   *    ** *   *   ** *   ** *   **  * *   ** ****    *   *   
** *  ** **** **  *  * **  **    * *  *   * ***  *  ** * *   ** ******   ** *   *   
 ****  ***  *   * *      ** ****   * **** ***
    **  ****   *  *   *   * *    * **  ** ** *** **    *     
* *** ** ** ****** * ***** ** * ****  *  * **  * ***
* *  *  *** *  *  *   *     *** *******        ****     ******* *** **  ****** ** *    *  *    * 
**    ***        ***  ** *   **  *  ***   * ** ****  **  * * *   ** *** **        *    *   *   *
   ******  ***  *   **  * *  *******  **  * ** *** *  **
    }
**** * **   *   ** **  **** *********  **
*  *  * *   ** * * ***  * **  *   ** **   * *   ** *** ** *   * *** * ** *    ***   
    else
 * ** *  * *  * ***    **  ** ** ** **  * *******   *********    * ** ** 
}
answered by (100 points)
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
Case 3: Wrong output
Case 4: Wrong output
Case 5: Wrong output
Case 6: Wrong output
Case 7: Wrong output
commented by (4.3k points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
Case 3: Correct output
Case 4: Correct output
Case 5: Correct output
Case 6: Correct output
Case 7: Correct output
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
#include <stdio.h>
#include <string.h>
int main ()
{
 ******* ***     *  * ****       str[1000]={0};
 ****   * *    *  ** * ** * j , n,sum1=0,sum2=0;
  *  *   *  *  *  *       * * *  ***  * * ** *  **
 * * ** *   *    **       = strlen(str);
***** *  **   * **** *     *   = 0;j < n;j++)
 *      * ***** *   * ** * 
*  * **  *  * ** *        ***  ****** * **  * *  ***** * == 0||j%2 == 0)
**        * * *   * ***      *   *  ***    *****  *  *     ** *  **** = sum2 + str[j] -48;
 ** ******* ***  *  **  ** *  * *  * ** **    * *  *** *** if ( j%2 != 0 )
 *** ** ***     **** ****  **    *  **  * * * * ** * * * ** **** * ***   * *  ****  = sum1 + str[j] -48;
***  * ******   ** *** **  
 *      *** * * ** *      ** * (sum1 > sum2)
*     * * * ** * * *  **   * **
*  ***  **  *** *    **   * * * *   * ***  * * ** ** - sum2) % 11 == 0 )
*** * *      ******** * ** * *** * * * *   ***    ****** *  ***  \n");
*    *  **  *** * * **   *** *  ** * ** **  **   ****  
  **       *** * **  * ** ** ** **    ***     ** *       **   \n");
**  ** *** **   *** *  *
    ** *   *   **   *  **   **   if(sum1 < sum2)
   **  *  * * * * **   ** *  *
 **** ** *** *     * *  ** * ***   * ****   * **  * - sum1) % 11 == 0 )
    **  **   **     *** * ****   * * * *** *  *  ** ****   *  \n");
 *** ** * * *    * * *** ** * *   * * *  *   *  
 * * **** ******  **  *   ***** ***   * * ***  ** ** **  *** \n");
  * ******* * ***  ******    
* * **   *   * *** * *** *  if (sum1 == sum2)
  * *  * **     * * * *** ** *       **  * * * * **   *   \n");
    **** ** **  **** * *****     *** * * ** * * *  * * *
 * *   ** ** ** **  *   * ***    **   0;
}
answered by (209 points)
edited by
0 0
prog.c: In function 'main':
prog.c:7:13: warning: format '%s' expects argument of type 'char *', but argument 2 has type 'char (*)[1000]' [-Wformat=]
     scanf("%s",&str);
             ^
prog.c:21:5: warning: implicit declaration of function 'system' [-Wimplicit-function-declaration]
     system ("pause");
     ^~~~~~
commented by (4.3k points)
0 0
prog.c: In function 'main':
prog.c:7:13: warning: format '%s' expects argument of type 'char *', but argument 2 has type 'char (*)[1000]' [-Wformat=]
     scanf("%s",&str);
             ^
commented by (4.3k points)
0 0
prog.c: In function 'main':
prog.c:7:5: warning: 'gets' is deprecated [-Wdeprecated-declarations]
     gets(str);
     ^~~~
In file included from prog.c:1:0:
/usr/include/stdio.h:640:14: note: declared here
 extern char *gets (char *__s) __wur __attribute_deprecated__;
              ^~~~
/tmp/ccwAQ4Bd.o: In function `main':
prog.c:(.text+0x3c): warning: the `gets' function is dangerous and should not be used.
commented by (4.3k points)
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
Case 3: Wrong output
Case 4: Wrong output
Case 5: Wrong output
Case 6: Wrong output
Case 7: Wrong output
commented by (4.3k points)
0 0
prog.c: In function 'main':
prog.c:7:13: warning: format '%s' expects argument of type 'char *', but argument 2 has type 'char (*)[1000]' [-Wformat=]
     scanf("%s",&str);
             ^
commented by (4.3k points)
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
Case 3: Wrong output
Case 4: Wrong output
Case 5: Wrong output
Case 6: Wrong output
Case 7: Wrong output
commented by (4.3k points)
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
Case 3: Wrong output
Case 4: Wrong output
Case 5: Wrong output
Case 6: Wrong output
Case 7: Wrong output
commented by (4.3k points)
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
Case 3: Wrong output
Case 4: Wrong output
Case 5: Wrong output
Case 6: Wrong output
Case 7: Wrong output
commented by (4.3k points)
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
Case 3: Wrong output
Case 4: Wrong output
Case 5: Wrong output
Case 6: Wrong output
Case 7: Wrong output
commented by (4.3k points)
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
Case 3: Wrong output
Case 4: Wrong output
Case 5: Wrong output
Case 6: Wrong output
Case 7: Wrong output
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
#include<stdio.h>

int main()
{
*  ** * * ****  * * *    arr[1000];
* *** *  *  ***     * ****   even=0,odd=0,i,n,sum=0;
   
*  * * *  *  *** ***  ** *   *******
   
***   *** ** *  * *  * ***  *** ** * 
   * ***   * * * *  ***
   
**    **   **   **** *  ** * *** *   
** ** ** *      ****  
*   *********   ** *** ***** ** *    **   * *    
*  * *  * **        ** *      *  * ****** * * *  ***    *****  * *   * * *** ***** **  **
 ** * ** * **   ** *   * ** *** *  ****  **    *  if(i%2!=0)
*   * **   *  ****  **  * * **  * *  **  *****     ** ****   *   *   *  ****  ****  ** * * *    **  ***** *****  *       ***** * ** *  *** ***   
   *** ***  * ****    
 * *  * *  ** **** * * *** *  
   
   * *  ** **  *       * *** **   
*  * * **** **    *  **
*** *** * **   ** *  *       **         *    * *    *   ****** ***  *   *      *  * **** **    **   ******  *******    *   ***   
  ***     ***      **
  **   * *****  * * * *  if(sum%11!=0)
******** * *  *  
** ** **  **** * * * **  *  ** * **   *  ** *** *****     **  * ***  ** 
 *  * **** *  *  *   ***
   
  
   ****   *** ***** **  *   
  *    * **** *  ***  ****  *   
}
answered by (174 points)
0 0
prog.c: In function 'main':
prog.c:8:12: warning: format '%s' expects argument of type 'char *', but argument 2 has type 'int *' [-Wformat=]
    scanf("%s",arr);
            ^
commented by (4.3k points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
Case 3: Correct output
Case 4: Correct output
Case 5: Correct output
Case 6: Correct output
Case 7: Correct output
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
#include<stdio.h>

int main(){
***    *  *  **   * * **  ***  one=0,two=0,i,ans;
**   * *       **     **   ** * ch[1000];
* * * *   * ** * *****  *  * * * * * **    *    ** *
    
*  **        *  **   ** ** **  *  *** 
 *  ** ** * **  **   * *** * **    *** * *  *           
***       *** *** ***  * ** * *     *  * ***   ***     * *****  * ** ** *  ***
****     **     *** ***   * *    * *   * *** * **   
*  **    ** **** *** *  *  ***    * ** ** *   * * *** 
**   *    ** **   ** * *  *  ** * * ***  *******  **    *** ** **    *    ** *
** ** **    * ** *    ** *    *** * **     **** **   *
    }
    
* ** * ****  *    * *  **  ** **
    
*  ** * * * *  *    ** ** * *** 
** * * * *  **   * * *    ***** *  * *   * *   *
*  ***  **   **   ***       * **   **  
*  ** ***   *******  ** *  ********  ** **  * **** * 
}
answered by (192 points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
Case 3: Correct output
Case 4: Correct output
Case 5: Correct output
Case 6: Correct output
Case 7: Correct output
commented by (4.3k points)
Page:
Welcome to Peer-Interaction Programming Learning System (PIPLS) LTLab, National DongHwa University
English 中文 Tiếng Việt
IP:172.70.127.4
©2016-2024

Related questions

2 like 0 dislike
5 answers
第二次期中考補考-第五題
[Exam] asked Dec 23, 2017 in 2017-1 程式設計(一)AD by 楊修俊 (30k points)
ID: 38942 - Available when: 2017-12-23 09:00 - Due to: 2017-12-23 12:10 | 824 views
2 like 0 dislike
18 answers
第二次期中考補考-第四題
[Exam] asked Dec 23, 2017 in 2017-1 程式設計(一)AD by 楊修俊 (30k points)
ID: 38941 - Available when: 2017-12-23 09:00 - Due to: 2017-12-23 12:10 | 2.2k views
4 like 0 dislike
16 answers
第二次期中考補考-第三題
[Exam] asked Dec 23, 2017 in 2017-1 程式設計(一)AD by 楊修俊 (30k points)
ID: 38940 - Available when: 2017-12-23 09:00 - Due to: 2017-12-23 12:10 | 3.8k views
3 like 0 dislike
12 answers
第二次期中考補考-第一題
[Exam] asked Dec 23, 2017 in 2017-1 程式設計(一)AD by 楊修俊 (30k points)
ID: 38938 - Available when: 2017-12-23 09:00 - Due to: 2017-12-23 12:10 | 1.9k views
2 like 0 dislike
3 answers
第二次期中考補考題庫 -10 唐．喬凡尼
[Normal] Coding (C) - asked Dec 20, 2017 in 2017-1 程式設計(一)AD by 楊修俊 (30k points)
ID: 38168 - Available when: Unlimited - Due to: Unlimited | 468 views
12,783 questions
183,443 answers
172,219 comments
4,824 users