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

0 like 0 dislike
11.8k 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 | 11.8k 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:172.71.254.247
©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 | 8k 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 | 9.5k views
12,783 questions
183,442 answers
172,219 comments
4,824 users