0 like 0 dislike
17.6k 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 | 17.6k 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:172.69.17.66
©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
| 12.3k 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
| 14.7k views
12,783 questions
183,442 answers
172,219 comments
4,824 users