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

0 like 0 dislike
25.7k 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 | 25.7k 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:104.23.243.6
©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 | 17.6k 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 | 21.9k views
12,783 questions
183,442 answers
172,219 comments
4,824 users