0 like 0 dislike
11.9k 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 | 11.9k 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:172.69.17.136
©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
| 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
| 9.5k views
12,783 questions
183,442 answers
172,219 comments
4,824 users