0 like 0 dislike
22.8k 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 | 22.8k 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:104.23.243.165
©2016-2026

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
| 15.6k 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
| 19.3k views
12,783 questions
183,442 answers
172,219 comments
4,824 users