CareerCup

Understandable version of code.

#include<stdio.h>
#include<stdlib.h>

void  print_multiple_of_all_ones(int num)
{
    int count=1,rem=1;

    while(rem)
    {
        rem = ( rem*10 + 1 ) % num;
        count++;
    }

    while(count--)
        printf("1");

    printf("\n");
}

int main()
{
    int num;
    scanf("%d",&num);
    print_multiple_of_all_ones(num);
    return 0;
}

#include<stdio.h>
#include<conio.h>
void main()
{ clrscr();
int n, num=1, r, x=1;
scanf("%d",&n);
while(x)
{ r=num%n;
if(r==0)
x=0;
else
{
num=r*10 +1;
}
printf("1");
}
getch();
}

sorry i forgot to check for unit digit '3'.. bt d logic is clear from d code..

answer in Java:

public static  void printAllOnes(int n)
	{
		System.out.println("Input: "+n);
		if(n%10!=3) {
			System.out.println("Output: Error !");
			return;
			}
		else {
			boolean cont = true;
			int num = 0;
			int length = 0;
			// get integer consisting only 1's
			while(cont){
				if(num>0 && num%n==0) 
				{
					cont = false;
					System.out.println("Output:"+num);
				}
				else{
					num = (int) (num + Math.pow(10, length));
					length ++;
					}
			}
		}
	}

#include<stdio.h>
#include<conio.h>
#include "stdlib.h"




void main()
{
int n, num=1, r, x=1;
scanf("%d",&n);
while(x)
{ r=num%n;
if(r==0)
x=0;
else
{
num=num*10 +1;
}
if(num>65535)
{
printf("integer overflow");
return;
}
}
printf("\t %d",num);
}

#include<stdio.h>

int main(){
int count=0;
int number,k;
scanf("%d",&number);
int temp = number;
printf("number = %d \n",number);
while(temp != 0){
temp = temp/10;
count++;
}

int start = 0;
for(k=0;k<count;k++){
start = start*10+1;
}
start = start*10+1;
printf("the start value is %d \n",start);
while(start != 0){
start = start*10+1;
count++;
start = start%number;

}
for(k=0;k<=count;k++){
printf("1");
}
printf("\n");
return 0;
}

Hi,

As per my understanding of the problem we need to print multiples of the number ending in 3 which consists of all 1's.But how is 111(7) a multiple of 3 and 111111(63) a multiple of 13.Can somebody elaborate the question please.
Thanks in advance.

public void test() {
System.out.println(bar(3));
System.out.println(bar(13));
System.out.println(bar(23));
}

private String bar(int x){
int i = 0;
int rem = 0;
StringBuilder sb = new StringBuilder();
do{
rem += foo(i, x);
i++;
sb.append("1");
} while (rem % x !=0);
return sb.toString();
}

private int foo(int power, int mod){
int retVal = 1;
for(int i =0; i<power; i++){
retVal = (retVal*10) % mod;
}
return retVal;
}

/*This code properly maintains overflow by returning the number of 1's in the answer */
int f(int num)
{
    int a = 111;
    int length=3;
    
    while(1)
    {
        if(a%num == 0 ) return length;
        a = a*10 + 1;
        if(a>=num) a%=num;
        length++;
        
    }
}

int main()
{
    int length = f(23);
    int i;
    for(i=1; i<=length; i++)
    printf("1");
    printf("\n");
    return 0;
}

#include <iostream>
using namespace std;

int main() {
	// your code goes here
	int t;
	cin>>t;
	while(t--){
		long long n;
		cin>>n;
		long long cur = 1;
		while(cur%n!=0){
			cur = cur*10+1;
		}
		cout<<cur<<endl;
	}
	return 0;
}