0 like 0 dislike
19.8k 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 | 19.8k 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:172.69.7.60
©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
| 13.8k 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
| 16.5k views
12,783 questions
183,442 answers
172,219 comments
4,824 users