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

0 like 0 dislike
18.7k 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 | 18.7k 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:172.69.6.119
©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 | 12.9k 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 | 15.4k views
12,783 questions
183,442 answers
172,219 comments
4,824 users