Yoda nzec

[dursalakihoegalo]

Please tell me how to take input in YODA? I am taking input like this:

for i in sys.stdin:

[numerix]

That is a correct way.

[dursalakihoegalo]

it actually is working in ideone. are there any test cases which yield NZEC?

[dursalakihoegalo]

input

65r865ghcvhgc
hgcgcgcv!@#$#54
vmhvbvhv..564!!#jhghj
hgcvhgv!$THkl

my output

0
0
0
1
0

[numerix]

My AC solution produces the same output.
I had no problems with NZEC. Just read the input using stdin and process it correctly.

[dursalakihoegalo]

I tried all test cases I can think of but still NZEC.

[dursalakihoegalo]

I increased the limit of factorial calculation range to 100 ans got a WA.
Now what to do?

[dursalakihoegalo]

Please suggest some test cases that I may go wrong.

does xpspxs answer 6?

[numerix]

Please suggest some test cases that I may go wrong.

That's your task. If you post some test cases, I can check the output with my AC solution.

does xpspxs answer 6?

Yes.

[dursalakihoegalo]

I found in the comments the answer of "aa11Bbb" is 6. Is that correct? I am getting 2.

[numerix]

I found in the comments the answer of "aa11Bbb" is 6. Is that correct? I am getting 2.

Your answer is correct.

[dursalakihoegalo]

would you like to see my code?

[dursalakihoegalo]

I have done many tests but it still gives WA...
my code:

EDIT: AC

[numerix]

Try this one:

xxxxx

[dursalakihoegalo]

Woowooooooooo.....AC..
You are great numerix. After you retire from SPOJ, please leave some great tips for us would be python coders.

[ankitrocks]

my code gives right answers for all possible test cases i know but still wa why i dont know

[daft_wullie]

condolences.
If you would like to get some hints, you should provide some information, e.g. description of your algorithm.

[ankitrocks]

i am using this

tot is total number of characters/2;
and num[i] stores occurences of characters between a and z

int fac[tot+10];
fac[0]=fac[1]=1;
for(i=2;i<=tot;i++)
fac[i]=i;
for(i=0;i<26;i++)
{
if(num[i]>1)
{
for(j=2;j<=num[i];j++)
{
for(k=j;k<=tot;k=k+j)
{
if(fac[k]%j==0)
{
fac[k]=k/j;
break;
}
}
}
}
}
unsigned long long ans=1;
for(i=2;i<=tot;i++)
ans=ans*fac[i];
cout<

[daft_wullie]

Perhaps you should have asked your question in another part of the forum. This area is python related.