PROBLEM LINK:
Practice
Author: ADMIN
Editorialist: SUSHANT AGARWAL
DIFFICULTY:
CAKEWALK
PREREQUISITES:
Basic looping,Basic Input/Output
PROBLEM:
Rewrite small numbers from input to output. Stop processing input after reading in the number 42. All numbers at input are integers of one or two digits.
EXPLANATION:
Please refer to the sample solution given by editorialist.
EDITORIALIST’S SOLUTION:
Editorialist’s solution can be found here.
3 Likes
#include<iostream.h>
void main()
{
int a,flag=0;
do
{
cin>>a;
if(a<=99 && a>=(-99) && a!=42)
{
flag=1;
cout<<a;
}
}while(flag==1);
}
2 Likes
import java.util.*;
public class inToOut{
public static void main(String [] args)
{
Scanner in = new Scanner(System.in);
while(in.hasNextInt() )
{
int temp = in.nextInt();
if(temp == 42)
break;
System.out.println(temp);
}
}
}
1 Like
#include<stdio.h>
int main()
{int i,b[100];
for(i=0;i<100;i++)
{scanf("%d\n",&b[i]);
if(b[i]==42)
break;
else
printf("%d\n",b[i]);
}
return 0;}
#include<stdio.h>
int main()
{
int a=1;
while(a>=(-99) && a<=99)
{
scanf("%d",&a);
if(a!=42)
printf("%d\n",a);
else
break;
}
return 0;
}
1 Like
#include<stdio.h>
#include<conio.h>
void main()
{
int x;
printf("\n Enter a two digit no to stop enter 42 “);
scanf(”%d",&x);
if((x>99)||(x=42))
break;
else
printf(“the entered no is : %d”,x);
}
1 Like
#include<stdio.h>
main()
{
int num=0;
while(num<100)
{
scanf("%d",&num);
if(num!=42)
printf("%d",num);
else
continue;
}
/this is corect answer even tho sowing false/
#include <stdio.h>
int main ()
{
int n;
for (;n!=42;)
{
scanf ("%d",&n);
printf ("%d\n",n);}
return 0; }
import java.util.Scanner;
public class LifeTheUniverseAndEverything {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
while (true) {
int i = scanner.nextInt();
if (i == 42)
break;
System.out.println(i);
}
}
}
#include<stdio.h>
int main(void)
{
unsigned int num;
for(;printf("%d",scanf("%d",&num)!=42););
return(0);
}
import java.util.Scanner;
/**
*
}
void main()
{
int i,a[6];
a[6]={1,42,84,7,8,9};
for(i=0;i<6;i++)
{
if(a[i]<=a[i+1])
{printf("%d",a[i]);
printf("\n");
}else
break;
}
getch();
}
#include<iostream.h>
using namespace std;
main()
{
int n;
cout<<“enter a number”;
cin>>n;
while(n>-99 && n<99 && n!=42)
{
cout<<“enter a no.”
cin>>n;
cout>>n;
}
1 Like
#include<stdio.h>
#include<conio.h>
void main()
{
int n;
while(1)
{
printf(“Enter the number:”);
scanf("%d",&n);
if(n==42)
break;
printf("%d",&n);
}
}
my program is having a compile error:
package codechef;
import java.util.Scanner;
public class chef {
private static Scanner input;
public static void main(String[] args) {
input = new Scanner(System.in);
int value = 0 ;
do{
System.out.println("Enter a number: ");
value = input.nextInt();
System.out.println(value);
}
while(value != 42);
System.out.println("Program terminated");
}
}
package package1;
import java.util.Scanner;
public class Class1 {
public static void main(String[] args) throws Exception {
while(true)
{
System.out.println("Enter an integer");
Scanner sc = new Scanner(System.in);
int num = sc.nextInt();
if( (num<100) && (num>(-100)) && (num!=42) )
System.out.println(num);
else
return;
}
}
}
include
using namespace std;
int main()
{
int i;
do
{
cout << "Enter i : ";
cin >> i;
if (i==42 && i<100 && i>-100)
{
break;
}
else
{
cout << i;
}
} while (i != 42 && i<100 && i>-100);
return 0;
}
#include< iostream >
#include< conio.h >
using namespace std;
int main()
{
int a[1000],i;
cout <<“enter integers” << endl;
for( i=1 ; i<=1000 ; i++ )
{
cin >> a[i];
if ( a[i]==42 )
{
break;
}
}
getch ();
return 0;
}
#include<stdio.h>
main()
{
int a[100],i;
printf(“enter numbers and enter 42 to end”);
for(i=0;i<100;i++)
{scanf("%d",&a[i]);if(a[i]==42)break;else printf("%d",a[i]);}
}
//C Solution by satadru97
#include<stdio.h>
int main(void)
{
int n;
while(1)
{
scanf("%d",&n);
if(n!=42)
printf("%d\n",n);
else
break;
}
return 0;
}