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

0 like 0 dislike
21.4k 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 | 21.4k 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:172.71.254.104
©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 | 14.8k 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 | 17.9k views
12,783 questions
183,442 answers
172,219 comments
4,824 users