0 like 0 dislike
20.8k 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 | 20.8k 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
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
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
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
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
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
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
0 0
Nice haha
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
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
Welcome to Peer-Interaction Programming Learning System (PIPLS) LTLab, National DongHwa University
English 中文 Tiếng Việt
IP:172.70.50.51
©2016-2025

Related questions

0 like 0 dislike
72 answers
[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
| 14.4k views
0 like 0 dislike
93 answers
[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
| 17.4k views
12,783 questions
183,442 answers
172,219 comments
4,824 users