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

0 like 0 dislike
17.1k 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 | 17.1k 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:172.69.59.97
©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 | 11.8k 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 | 14.1k views
12,783 questions
183,442 answers
172,219 comments
4,824 users