10070 - Leap Year or Not Leap Year and ...

[haaaz]

This is my source code for problem 10070. When I ran it on my computer, it worked well when I input a long number. But I got "Output Limit Exceeded" when I submit it. Can anyone help me?
[c]#include <stdio.h>
#include <string.h>
main()
{
char s[10000];
int y[10000];
//...
[/c]

Maybe 10000 elements are not enough?

[shamim]

I have submitted the following code for judging

i realize that i have to use strings to take input and i have used a technique to find the divisibility of large numbers by smaller ones. yet i get WA. Could someone help me to spot the mistake.


Cut :

[PdR]

[c]
if (leap)
{
printf ("This is leap year.\n");
if (isdiv (line, 11)) bulu = 1;
}
[/c]

Bulukulu festival (Happens on years divisible by 55 provided that is also a leap year)

I supose this is your problem

[tat tvam asi]

helo
... plus a small typo :
use the word "bulukulu" - as in the
prob. statement - not "buluculu" .
bye

[soyoja]

Dear haaz.
10000 elements array would be enough to store test data.
My program also use 10000 elements char array.
Finally, I got "Accepted". ^^

[Almost Human]

I need some suggestion here ...

Do I have to use very long number in this case ???
Can I just use int or long ???

please help!

[shahriar_manzoor]

Yes numbers are out of range of any conventional data types - integer, long, int64 or long long etc

[titid_gede]

you must use big integer, pak andoko. anyway i do not know how is the upper limit.

[Almost Human]

Thanks everyone..

I still don't know the precise limit for this...

but I used my verylonginteger class and used 2000 for MAXDIGIT and got AC for that ...

[awik_10]

what's wromg with my program ? would anyone tell me what's my wrong? this is my code :
[c]
#include<stdio.h>
#include<string.h>
#include<ctype.h>
#include<stdlib.h>

int main()
{
char s[10000];
int y[10000];
int len,i,m4,m15,m55,l,h,b,a,cek;
while(scanf("%s",s)==1)
{
fflush(stdin);
len=strlen(s);
for(i=0;i<len;i++)
{
if(!isdigit(s)) exit(1);
}
if(len<3) continue;
if(len==4)
{
cek=0;
for(i=0;i<4;i++)
{
cek=cek*10+s-'0';
}
if(cek<2000) continue;
}
if(len==0||s[0]<48||s[0]>57)
break;
for(i=9999;i>=len;i--)
y=0;
for(i=len-1;i>=0;i--)
if(s[len-1-i]>=48&&s[len-1-i]<=57)
y=s[len-1-i]-48;
m4=(y[1]*10+y[0])%4;
m15=y[len-1];
for(i=len-2;i>=0;i--)
m15=(m15*10+y)%15;
m55=y[len-1];
for(i=len-2;i>=0;i--)
m55=(m55*10+y)%55;
l=h=b=0;
if(m4==0)
l=1;
if(m15==0)
h=1;
if(l==1&&m55==0)
b=1;
if(l==1)
printf("This is leap year.\n");
if(h==1)
printf("This is huluculu festival year.\n");
if(b==1)
printf("This is bulukulu festival year.\n");
if(l+h+b==0)
printf("This is an ordinary year.\n");
printf("\n");
}
return 0;
}[/c]

thank you
best regards,
awik

[shamim]

Your code looks ok, but there are few unnecessary things,

for(i=0;i<len;i++)
{
if(!isdigit(s)) exit(1);
}

there is no need to make this test,

One thing u could try is creating a single function to test for divisibility.

[awik_10]

One thing u could try is creating a single function to test for divisibility.

what's that mean, can you explain me for more detail, please?
i've removed the unnecessary thing from my program, but i still don' t get the acc.
thank you,
awik

[turuthok]

Awik, probably this problem gives us a number with very-long digits. Can somebody confirm this statement ???

-turuthok-

[turuthok]

Sorry, I should check for previous posts before making comments ...

Yes, this problem involves big-integer ...

-turuthok-

[dserrano]

[quote="10153EN"]I solved this problem by storing the year number (which can be very large) in an integer array. Then check this year with some number theory.

For example, if a numbe if divible by 3, then the sum of all of it's digits is ....[/quote]

Excellent idea!!
What's the rule for knowing if a number is divisible by 11?