10929 - You can say 11

All about problems in Volume 109. If there is a thread about your problem, please use it. If not, create one with its number in the subject.

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Raiyan Kamal
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Post by 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
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Post by sml »

Raiyan Kamal wrote: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
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10929--Algorithm

Post by 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!
i wanna give it a try....

dust_cover
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Post by dust_cover »

Hi I got the problem & got AC!
No need to explain!
Thanx Anyways!
:D
i wanna give it a try....

Waddle
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10929 - You can say 11

Post by 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
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Post by 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.
Ami ekhono shopno dekhi...
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Waddle
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~O~ I don't really understand what you say...

Post by Waddle »

The code has been removed.
Last edited by Waddle on Wed Jun 06, 2007 3:05 pm, edited 1 time in total.

Jan
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Post by 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

Code: Select all

system("pause"); 
Hope these help.
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Waddle
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Post by Waddle »

Thank you!!!!!!

hridoy
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Post by hridoy »

How this algorithm works?

Jan
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Post by 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.
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amine.hamdaoui
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Output Limit exceeded

Post by amine.hamdaoui »

Why do i have Output Limit exceeded?!!!!
[codeIt was accepted[/code]
Last edited by amine.hamdaoui on Sun Aug 12, 2007 10:57 pm, edited 1 time in total.

little joey
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Post by little joey »

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

Code: Select all

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

Code: Select all

if((line.size()==1)&&(line[0]=='0')) break;
you'll get accepted. Then, please, remove your code.
The biggest problem with most problems is not how to solve the problem, but how to not solve what is not the problem.

apurba
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where's wrong in the code?

Post by apurba »

Code: Select all

removed after ac

Code: Select all

keep dreaming...

Obaida
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Oh..........No...

Post by Obaida »

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

Code: Select all

removed
Last edited by Obaida on Wed Mar 18, 2009 6:26 am, edited 1 time in total.
try_try_try_try_&&&_try@try.com
This may be the address of success.

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