11-01 Exercise 1: Find out all prime numbers between n and m

0 like 0 dislike
24.9k views

Write a program that get the input of two integers n and m(n < m). Find out all prime numbers between n and m (not include n and m), then display them on the screen.

寫一個程式 輸入兩個整數n和m (n < m)，找出所有n到m之間的質數(不包含n和m)。

Example input 1

1 10

Example output 1

2 3 5 7

 

Example input 2

10 20

Example output 2

11 13 17 19

 

[Exercise] Coding (C) - asked in Chapter 6: Loops by (5.2k points)
ID: 29829 - Available when: 2017-11-02 18:00 - Due to: Unlimited
edited by | 24.9k views

99 Answers

0 like 0 dislike 
Hidden content!
#include<stdio.h>
int main(){
    int a,b,c=0;
   * *    * *  **   * ******* **** * ** * * %d",&a ,&b );

*** * ***   ** * * *  * * * i=a;i<b;i++){
  ***    ** ******* * **** **** ** ** **   * * **    ** j=2;j<=b;j++){
*  ***** **  * ** ** * ***  * *   *   * **     ** ******** * * *   ****        *** && i%j==0) break;
  **   ** *   **  ***  *** * * **  *    ***   ***   **  ** *       *   *** * if(i==j){
   *** * *****   **     **  * * **     *  *** *    *** *  *** *    *       *  ******* **   *    * *** * *
*      ** ***  **   *    ** *** * *** *   * ** **  ** ** ** *   *   * * ***** * *** ***    ** ***  ****    * *     ***  *** *   * *  ** *** * ****     ** ****
*  * ** ***    *  **  * ***       ** *** *** ** *** ***   ** *    * ** ** *    *****  ** * *  *** * *   **      *   **  **               
*     ***    *  ****  * ** ** * ** * * **** ** ** **** **  *  *  **  *  *  ** * * ** *    *  * ***   **  **    *
  ***   ****  **    **  * * **    *** **    *** * *  **  * * * * *        * **   * * **  * ***  **   *  ** * **** *  printf(" %d",i);
     * * ****   *     ***** ***     *** **  ***     ******* * * ** ** * 
 *  *   ***     ** ** *  *     ********* * * **  ** 

    }

    return 0;
}
answered by (-116 points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
Case 3: Wrong output
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
*  *     ** ** *    *** 
 ** * ***   **   
 **   ** ******  *  *

int main()
{
int i,n,m;
** ****   *  ****   ***** * *
***    * *  **** * **** * * **
***    * **  *** *  

{



*   ****  *   ** **
{
*     
**  **
}
 * *  * *   * *

    *


   *** * *  *   ** ***


}}
*  *  0;
}
answered by (-336 points)
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
Case 3: Wrong output
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
 *      ******   *    
*  *    *  ** *   ** *
 *****   ***   * *  * 

int main()
{
int i,n,m;
** *  *   *  ** ****  * *
** **  * *  ** *        * 
 **   *   * ***  ******  

{



  **   * *  ********  
{
 *   **
 * **
}
** ** * ** * * 

  ** 


 *  *  * ** * * *    * ***


}}
**   0;
}
answered by (-336 points)
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
Case 3: Wrong output
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
#include<stdio.h>
int main(){
    int a,b,c=0;
*  **     *****  ****   *** * ***  *   *   ** %d",&a ,&b );

 **  * * *  **** *  **** *    i=a;i<=b;i++){
   **** *     **      * * *  * *   *    ** *  **  **    **  * *   j=2;j<=b;j++){
***    ***** *  ****   **** ***  *** *   *   *        ****** * ***     * **** ******* && i%j==0) break;
* *    **  * ***  *  *  ** *  ****  **  * * *****   *   ******** * *  * **** if(i==j){
  ** * **  **** *****     * * ** * ****    ** * ****  **  * *      ** *   *   **  ***    **  *       *****   * ** ***
*   **    *    *   *  ***  ***   **  ** **** *     **  **   *  * **       *  *  ** *** *  * *  ** * *      * *** *   *     *  *  **  ** *  * **  *  *   * *
***  *  *  **    *    *   **  *   * ****  ** **  * *     ** **    * * **  ** **   * * * **** * * ***   *  ** **  * **  *** * ** *** *** *****  
**  **** * ** * *    ******  *** ***    **** **  **    **  *  *   **        *  *  **     ******  ****** *  
   ** **** *  ***  ******  *   **  ****    * *    * *   ****    *** ** *******   *  * * * ***  * printf(" %d",i);
**      *** *      ** ** *    *      ***  * *   ** *  *** *  *** **  * * *  *  *   
 ***** *   ****  **  ** *** *   ****** **         **

    }

    return 0;
}
answered by (-116 points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Wrong output
Case 3: Wrong output
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
#include<stdio.h>
int main(){
 ***  ***  * ** ** ****** *  ** ** a,b;
***    * *  **   *   **  ******* ** * ** * %d",&a ,&b );

* * * *  **  * * **** *** **   * i=a;i<=b;i++){
 *** ***  *  ** *      * * *  **** *  *  *** ***** * **   j=2;j<=b;j++){
 * * ** * *  ***  **    **   **  * **  *      **     * ** * ***   * ** ***   && i%j==0) break;
 * ***  *  * **   **** ****** * ** * *******  *   *** *      *  * **  * *   *  ** * if(i==j){
  **   *  *  * *** * ****** ***   **** ***   ** * ** *  *** *  *  * ******  *  ***   *  *** **    *****  ****  *  **  ** ",i);
*** ***  *    **** * ***    *   *** * *    ** * **   **    * ** * **  * **** **  *
  **** *  *  **  ** ** *  ** ** * ****     * **  * ***  

** *  *** * **  ** * ** *
** *    **    *   *   *  *  * * 0;
}
answered by (-116 points)
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
Case 3: Wrong output
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
 ** * ******* **
        ** **    **
int main(){
   ***** ** *  * *** * **  ** a,b;
   ** *  *  ** ***  **  *  * * ** ** *  *** * ***   ***** ,&b );

   **    *    *** ******  *  * * i=a;i<=b;i++){
 **  * * **  *   **    ** ** **      **    * **  **    ** *  j=2;j<=b;j++){
* ** *  * *** ** *    **   ** ** **  *     *****  *   ** * * ** *  ** * * *** ** *** && i%j==0) break;
   ********  * *  ****            **   *  * * **** ** * **    ** *      **     *** if(i==j) printf("%d ",i);
*   *   ** * *  *  * ***  ** **** * **  *   ** * *  
  **    **** * * * *
* * *** **  ** * ** * *  **  * ** 0;
}
answered by (-116 points)
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
Case 3: Wrong output
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
#include <stdio.h>
#include <stdlib.h>

int main()
{
   *   *** * *   *  *   m, n, divider, prime[100], cnt=0, i=0;

**** ***  * ** ** ** ***  **  **** ** %d", &m, &n);

** **  * *   * ** *   ** *  * = m + 1; m < n; m++)
*     * ***** **   ***     **** *  
*        ** **  ** * ** ******* * **** *  **  ** ** *   **** = 2; divider < m; divider++)
*  **      ***  **** * ** **** **  *   *   ** * *
  **  **  *** * * ***  * **   * *  * * **  ** * **  * **  * * ** *  ** **    *** % divider == 0)
**  ** * *     ** ***  *** **     *   *  **  **  *  ****  **     *  *  ** *  *   *  * *  **  *    ** 
*  *** *** **   *   ***    *     *    *  *   ** **
    **  **  ****    **   *  * * *   * * ** **  ******   *    == divider)
  * * *     ***   **   *   *  *  ** ****** * *
****      ***  ******  *****  *** *****  **** ** ** **** *  *    **  ** ****    **      **** = m;
 ****** *  * *   *   ***  *****   *  *   * ** *   **
** ***  *       *    ** ** 
  *** * *  *  *** ***** ***** **   i<cnt;i++)
 **  ** ** ** *  *  ****** 
* *  *  *** **** ****  *      *  ***     **  * ** ***   *** ** * * ****   prime[i]);
* *   *** * *    *      ** *****   * * ****      *   * +1 < cnt)
   *** ** *     *   *  **  **** ***** *** * *  **   *****    *   * **  **  ** *  *   ** *** ");
    }

   ** ****   * ****   *** * * * 0;
}
answered by (5.2k points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
Case 3: Correct output
commented by (4.3k points)
0 0
Nice haha
commented by (12.1k points)
0 like 0 dislike 
Hidden content!
#include <stdio.h>
int main()
{
    int n,m,i,j,k;
    scanf("%d %d",&n,&m);
****      ** *   * * *  *  ** **   *** ****  **
**  * **  **  *    *   *     ** *   **   ** ** ***** *   *  ** 
 *  *** ***  **** ** *  *** *  ***  **  ** ***** *    *  *  ** *  **** ** ****
***   **  *   **   * *** *** * * *  **        ***    *   * *   **** *   ****  *  **   ****  **      *  **  *** ** *
  *    * * * *  *   * ** *  * * ***  *  *  **  * *  *  **   ***       * * *  ***   *  * *
* **   * **** *  * **    ***     *  *  ****   * *****  *** * ** ** * *** *** * * *  * * ****  * **    ** * ** * ",i);
*  **    *** * *   * *  **** ** *   **     ***  ** **   *      *         *      ****   *              *** **  ** 
**  * * ***  ** ***  **       **  *       *** ** **  **   * * ** **  ***   *  
 **  ** *   **   *     ** *****   ****** ****  * ** *  ** *  **   * **  **     ** if(k==0){
* **  ***   *   ** *  ***  ** *   *** * *  ***  * *  * ** **  *   *   ***  *    *  *** * **  *  * *  * ***   * * 
**   *   *  ***** ****  *  **        ***  *     *****      * *  **  *  *
** * ***  **  *  * **  *   *  *    ***  * **   * * *  
    }
    return 0;
}
answered by (54 points)
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
Case 3: Wrong output
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
#include <stdio.h>
int main(){
int i, a , b;
int flag;
scanf("%d    ** * **    ,&b);

for * *****   ***** *
{
* *  ***       * * **   * * *     *   * is testing\n",i);
* *** ***** ** **   **    **  j=0;
 **** * **  **    ***** * *   = 1;
 *  *  *** **  *  ******  * *   * * (j=2;j<i;j++)
  ** ***  **     **** *   * *
 * *     *****  **  *  *** *  ** *  * * ******* * * * * * **  *     ** *    %d is mode\n",j);
* **** **  ** ****  ** *** ** ***** * **   ** *  ** *  ** reminder=i%j;
 *  ****   ** **** ******    **  ** *   * ** *    *** (reminder == 0)
 ******  * *  ** ***** ****** * *  *  * * * * ** *  *  *   
***   *  * ******* **** * * ****  * **    **  *   *      *       ** *  * * * *   *  *** ***  * **** * **   **  **** **  **   * **
 * ** ***  ** * *    *  ***   ***   *   *    **  **   * ** * ****** *  ** **  **
 ** *  **   * * ** ** ** *     ***** **         ** **** *  *** * *   **  ***           
      * *   *  * *    **    *** *   * ** *  ***  * *** *
       ********   ** ** *
**  ****   * *****   ** *  (flag==1)
 *  ** *  ** * ** * *    *  ***   **  * ** ** * ** *  *  *  ** *   *     ",i);
}

return 0;
}
answered by (-141 points)
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
Case 3: Wrong 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:104.23.197.94
©2016-2026

Related questions

0 like 0 dislike
72 answers
11-01 Exercise 2: Find out all the factors of N
[Exercise] Coding (C) - asked Nov 2, 2017 in Chapter 6: Loops by semicolon (5.2k points)
ID: 29831 - Available when: 2017-11-02 18:00 - Due to: Unlimited | 17k views
0 like 0 dislike
93 answers
11-01 Exercise 3: Fibonacci numbers
[Exercise] Coding (C) - asked Nov 2, 2017 in Chapter 6: Loops by semicolon (5.2k points)
ID: 29830 - Available when: 2017-11-02 18:00 - Due to: Unlimited | 21.2k views
12,783 questions
183,442 answers
172,219 comments
4,824 users