0 like 0 dislike
17.4k 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 | 17.4k 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:172.69.214.204
©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
| 12.1k 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
| 14.4k views
12,783 questions
183,442 answers
172,219 comments
4,824 users