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

0 like 0 dislike
17.6k 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 | 17.6k 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:172.70.194.139
©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 | 12.3k 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 | 14.7k views
12,783 questions
183,442 answers
172,219 comments
4,824 users