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

0 like 0 dislike
11.5k 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 | 11.5k 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:172.70.131.84
©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 | 7.7k 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 | 9.2k views
12,783 questions
183,442 answers
172,219 comments
4,824 users