10929 - You can say 11

[rimu]

i'm very confused getting WA

can anybody post some critical i/o

my logic is as follows in brief:


gets( r );

for ( i = s = 0; r; ++i )
{
s = s * 10 + r - 48;
s = s % 11;
}

if ( !s ) then multiple

Thanks in advance for replying

[arif_pasha]

sorry above reply was mine..

[Jan]

I think there are no inputs like above.

I used gets() to take inputs. But There can be leading zeroes in the input.

Input:

Code: Select all

011
11
122321
0123123
0

So, change your code.

Code: Select all

....
    if(r[0]=='0')
        break;
....

The code above is wrong. You should use this...

Code: Select all

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

Hope it works.

[TISARKER]

According to jans Idea
Try this input

Code: Select all

00000000

what is the correct output for above input.?

[iishaque]

Plz give some critical input and output. Can any one tell Is rimu's algorithm right or wrong??
Plz help!!!!!

[SRX]

Plz give some critical input and output. Can any one tell Is rimu's algorithm right or wrong??
Plz help!!!!!

Code: Select all

     just count num's odd digits sum - num's even digits sum 
     then check it%11 
     then use Jan's way ( thanks  :D ) if(r[0]=='0'&&r[1]=='\0') break;  

[iishaque]

Thank you SRX for you help. I got AC.

[Martin Macko]

According to jans Idea
Try this input

Code: Select all

00000000

what is the correct output for above input.?

The 0000000 is not positive, so it may be only at the end of the input. But it's not because my AC recognized the end of the input by a single 0 in line.

[Schutzstaffel]

I'm getting WA and have no clue what it might be

This is my function:

Code: Select all

code ACC and thus removed :P

I sum the odd digits and subtract the even digits then check if the sum is divisible by 11. Did I miss something?
Thanks.

[Krzysztof Duleba]

I don't think that using atoi on a not null-terminated string is safe.

[Schutzstaffel]

You're right, it was because of that.
Thanks

[athlon19831]

I got Time Limit Exceeded,but i don't know why
my code:
#include "iostream.h"
#include "stdio.h"
#include "string.h"
#include "math.h"
int main(int argc, char* argv[])
{
char str[1010];
int s_odd,s_even;
int s;
int i,j;
while(cin>>str)
{
if(strcmp(str,"0")==0)
break;
else if(str[0]=='-')
break;
else
{
s_odd=0;
s_even=0;
for(i=0;i<strlen(str);i++)
{
if(i%2)
{
s_odd+=(int(str)-48);
}
else
{
s_even+=(int(str)-48);
}
}
s=int(fabs(s_odd-s_even));
if(s==0)
{
for(i=0;i<strlen(str);i++)
{
cout<<str;
}
cout<<" is a multiple of 11."<<endl;
}
else if(s%11==0)
{
for(i=0;i<strlen(str);i++)
{
cout<<str;
}
cout<<" is a multiple of 11."<<endl;
}
else
{
for(i=0;i<strlen(str);i++)
{
cout<<str;
}
cout<<" is not a multiple of 11."<<endl;
}
}
}
return 0;
}

[chunyi81]

You are calling strlen each time in the for loop you print the input number. And in fact, you can print out a char array directly just like C++ strings.

Just do a cout << str for the places where you print the input number character by character and the TLE should go away.

If not, you might wan to consider scanf/printf rather cin/cout.

Hope this helps.

[athlon19831]

You are calling strlen each time in the for loop you print the input number. And in fact, you can print out a char array directly just like C++ strings.

Just do a cout << str for the places where you print the input number character by character and the TLE should go away.

If not, you might wan to consider scanf/printf rather cin/cout.

Hope this helps.

Thanks you for your help. I got AC

[Zaspire]

Do not forget that you should output number with leading zeroes (if it has)