0 like 0 dislike
24.7k 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 | 24.7k 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:104.23.197.95
©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
| 16.8k 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
| 20.7k views
12,783 questions
183,442 answers
172,219 comments
4,824 users