10929 - You can say 11

[Raiyan Kamal]

If you just add/substract alternative digits and chek the final sum, then leading zeroes are not supposed to create any problem. Why are a lot of people so worried about the leading zeroes ?

[sml]

If you just add/substract alternative digits and chek the final sum, then leading zeroes are not supposed to create any problem. Why are a lot of people so worried about the leading zeroes ?

Because a lot of us were checking input[0] for '0', and exiting the program in the event that it was true. I had the same basic problem, though I was glad my algorithm was correct (without cheating! Not that there are many threads on this subject).

[dust_cover]

can someone tell me how the algorithm described above

(sum of odd digits-sum of even digits) works for 22?
It is a bit confusing for me!
--thnx in advance!

[dust_cover]

Hi I got the problem & got AC!
No need to explain!
Thanx Anyways!

[Waddle]

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
int sum1=0,sum2=0,k,i,j;
char a[1000];
while(1)
{
gets(a);
if(a[0]=='0'&& a[1]=='\0')
return 0;
k=strlen(a);

for(i=0;i<k;i=i+2)
sum1=sum1+a[i]-48;
for(j=1;j<k;j=j+2)
sum2=sum2+a[j]-48;

if((sum1-sum2)%11==0)
{
printf("%s is a multiple of 11.\n",a);
}
else
{
printf("%s is not a multiple of 11.\n",a);
}
}
return 0;
}

[Jan]

Search your problem first. Dont open a new thread if there is one already.

You have to initialize sum1 & sum2 for every case, not only for the first case.

[Waddle]

The code has been removed.

[Jan]

The problem states...

The given numbers can contain up to 1000 digits

So, to store it in an array of characters you need 1000 + 1 elements. The last '1' is for a NULL character. So, increase your array size.
And remove the following unnecessary line

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system("pause"); 

Hope these help.

[Waddle]

Thank you!!!!!!

[hridoy]

How this algorithm works?

[Jan]

It's a school time technique. An integer is divisible by 11 if the difference between 'the summation of the odd positioned digits' and 'the summation of the even positioned digits' is divisible by 11.

[amine.hamdaoui]

Why do i have Output Limit exceeded?!!!!
[codeIt was accepted[/code]

[little joey]

A string doesn't have to be zero-terminated. If you replace

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if(line[0]=='0' && line[1]=='\0') break; 

by

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if((line.size()==1)&&(line[0]=='0')) break;

you'll get accepted. Then, please, remove your code.

[apurba]

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removed after ac

[Obaida]

Some one please help me this shouldn't be happened I m getting WA....please help me...

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removed