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

0 like 0 dislike
14.4k 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 | 14.4k 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:172.70.194.178
©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 | 9.9k 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 | 12k views
12,783 questions
183,442 answers
172,219 comments
4,824 users