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

0 like 0 dislike
24.5k 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 | 24.5k 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:162.159.115.10
©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 | 16.7k 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 | 20.5k views
12,783 questions
183,442 answers
172,219 comments
4,824 users