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

0 like 0 dislike
15k 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 | 15k 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:172.69.59.23
©2016-2025

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 | 10.4k 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 | 12.4k views
12,783 questions
183,442 answers
172,219 comments
4,824 users