第二次期中考補考-第二題

3 like 0 dislike
1.4k 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 | 1.4k 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:172.70.130.82
©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 | 383 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 | 1.1k 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 | 1.8k 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 | 959 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 | 288 views
12,783 questions
183,443 answers
172,219 comments
4,824 users