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Re: 10929 - You can say 11

Posted: Sun Jul 07, 2013 7:40 pm
by Faithkeeper_Rangwan

Code: Select all

import java.util.*;
import java.math.*;
class Main 
{
	public static void main(String[] args)
	{
		BigInteger a,b;
		Scanner sc = new Scanner(System.in);
		a = sc.nextBigInteger();
		while(!a.equals(BigInteger.ZERO))
		{
			b =a.mod(BigInteger.valueOf(11));
			if(b.equals(BigInteger.ZERO)) System.out.println(a+" is a multiple of 11.");
			else System.out.println(a+" is not a multiple of 11.");
			a = sc.nextBigInteger();
		}
	}
}
Also got WA with this one, don't know how to fix this :(

Re: 10929 - You can say 11

Posted: Mon Jul 08, 2013 11:52 pm
by brianfry713
input:

Code: Select all

01
0
output should be

Code: Select all

01 is not a multiple of 11.

Re: 10929 - You can say 11

Posted: Thu Jul 11, 2013 4:40 pm
by Faithkeeper_Rangwan
brianfry713 wrote:input:

Code: Select all

01
0
output should be

Code: Select all

01 is not a multiple of 11.
AC'd

Thanks.

UVA 10929

Posted: Sun Oct 20, 2013 7:28 pm
by woaraka92
#include<stdio.h>
#include<string.h>

int main(){
char a[1002];
int i,sum;
while(1){
scanf("%s",a);
if(strcmp(a,"0")==0)
break;

sum=0;
for(i=0;a!='\0';i++){
sum=sum*10+a-48;
if(sum<11){
i++;
sum=sum*10+a-48;
}

if(sum>10)
sum=sum%11;
}

if(sum==0)
printf("%s is a multiple of 11.\n",a);
else
printf("%s is not a multiple of 11.\n",a);

}
return 0;
}

Re: UVA 10929

Posted: Sun Oct 20, 2013 7:33 pm
by woaraka92
whats the problem in this code?

Re: UVA 10929

Posted: Mon Oct 21, 2013 9:25 pm
by brianfry713
110 is a multiple of 11.

Re: UVA 10929

Posted: Wed Oct 23, 2013 9:18 pm
by woaraka92
Thanks brianfry713..

Re: 10929 - You can say 11

Posted: Thu Mar 13, 2014 10:00 pm
by jddantes
What's wrong with mine?
Also by positive number do they mean positive integers?

Code: Select all

#include <stdio.h>
#include <string.h>
int main()
{
    char number[1005];
    while(fgets(number, 1005, stdin)!=NULL)
    {
        number[strlen(number)-1] = 0;

        if(number[0] == '0' && number[1] == 0)
        {
            return 0;
        }

        int sum = 0;
        int i = 0;
        for(i=0; number[i];i++)
        {

            if(i%2 == 0)
            {
                sum+=number[i] - '0';
            }
            else
            {
                sum-=number[i] - '0';
            }

            //printf("%d %d\n",sum, number[i] - '0');
        }

        if (sum % 11 == 0)
        {
            printf("%s is a multiple of 11.\n",number);
        }
        else
        {
            printf("%s is not a multiple of 11.\n",number);
        }
    }

    return 0;
}

Re: 10929 - You can say 11

Posted: Fri Mar 14, 2014 9:02 pm
by brianfry713
That is AC code.
Yes they mean positive integers.

Re: 10929 - You can say 11

Posted: Sun Mar 16, 2014 1:22 pm
by jddantes
Why is mine WA?

Code: Select all

#include <stdio.h>
#include <string.h>
int main()
{
    char number[1005];
    while(fgets(number, 1005, stdin)!=NULL)
    {
        number[strlen(number)-1] = 0;

        if(number[0] == '0' && number[1] == 0)
        {
            return 0;
        }

        int sum = 0;
        int i = 0;
        for(i=0; number[i];i++)
        {

            if(i%2 == 0)
            {
                sum+=number[i] - '0';
            }
            else
            {
                sum-=number[i] - '0';
            }

            //printf("%d %d\n",sum, number[i] - '0');
        }

        if (sum % 11 == 0)
        {
            printf("%s is a multiple of 11.\n",number);
        }
        else
        {
            printf("%s is not a multiple of 11.\n",number);
        }
    }

    return 0;
}

Re: 10929 - You can say 11

Posted: Mon Mar 17, 2014 9:55 pm
by brianfry713
That is AC code.

Re: 10929 - You can say 11

Posted: Mon Mar 24, 2014 7:38 am
by uDebug
So, for this problem, it turns out that there are two outputs that are both accepted.

For, the following input

Code: Select all

            112233              
    00000000030800
   2937                           
                      323455693                 
          5038297          
          00000112234     
00112            
0
AC Output #1:

Code: Select all

112233 is a multiple of 11.
00000000030800 is a multiple of 11.
2937 is a multiple of 11.
323455693 is a multiple of 11.
5038297 is a multiple of 11.
00000112234 is not a multiple of 11.
00112 is not a multiple of 11.
AC Output #2:

Code: Select all

            112233               is a multiple of 11.
    00000000030800 is a multiple of 11.
   2937                            is a multiple of 11.
                      323455693                  is not a multiple of 11.
          5038297           is a multiple of 11.
          00000112234      is not a multiple of 11.
00112             is not a multiple of 11.

Re: 10929 - You can say 11

Posted: Tue Mar 25, 2014 12:15 am
by brianfry713
There probably aren't spaces in the judge's input.

Re: 10929 - You can say 11

Posted: Tue Mar 25, 2014 8:14 am
by uDebug
brianfry713 wrote:There probably aren't spaces in the judge's input.
That makes sense. Thanks.

Re: 10929 - You can say 11

Posted: Sun Mar 30, 2014 3:01 am
by jddantes
Why is mine not accepted?

Code: Select all

#include <stdio.h>
#include <string.h>
int main()
{
    char number[1005];
    while(fgets(number, 1005, stdin)!=NULL)
    {
        number[strlen(number)-1] = 0;

        if(number[0] == '0' && number[1] == 0)
        {
            return 0;
        }

        int sum = 0;
        int i = 0;
        for(i=0; number[i];i++)
        {

            if(i%2 == 0)
            {
                sum+=number[i] - '0';
            }
            else
            {
                sum-=number[i] - '0';
            }

            //printf("%d %d\n",sum, number[i] - '0');
        }

        if (sum % 11 == 0)
        {
            printf("%s is a multiple of 11.\n",number);
        }
        else
        {
            printf("%s is not a multiple of 11.\n",number);
        }
    }

    return 0;
}