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

0 like 0 dislike
19k 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 | 19k 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:172.69.17.67
©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 | 13.2k 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 | 15.9k views
12,783 questions
183,442 answers
172,219 comments
4,824 users