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

0 like 0 dislike
12.9k 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 | 12.9k 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:172.71.254.209
©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 | 8.7k 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 | 10.5k views
12,783 questions
183,442 answers
172,219 comments
4,824 users