What precision should we use in output?

Could I have WA#17 because of it?

Is double (long double) enough for us?

double is always enough for calculatings on Timus.

Then it (4 digits) should be written in russian statement.

To avoid precision problems it is possible to use only integer numbers in solving this task, but I have the same WA#17.

Do you use int64? Check it.
(50 000 * 50 000 = 2 500 000 000)

Yes, of cause, I use unsigned __int64.

I tried double, __in64 and long double and still WA17. That must be some other bug...

Maybe problem with the following code:

for ( i = 1; i <= m_N; ++i )
for ( j = 1; j <= m_N; ++j )
sum += t[ i ][ j ];

sum /= N * (N - 1);

It must be obviously changed to ...
for ( i = 1; i <= m_N; ++i )
for ( j = 1; j <= m_N; ++j )
sum += t[ i ][ j ]) / ( N * (N - 1));

overflow ...

Maybe it will be usefull :- )

And again I've tried both variants... Still WA17

Why both variants! First fragment implyies WA ( it`s obvious )
Did you convert N to __int64 ? It was noted that N*N > signed int.

Hmmm ... Send me your code ( see e-mail in my profile )

Yes. The bug what in these data types. I think I should love __int64 from this task and further. Thank you very much! I got AC.

I used __int64 (signed) to get total sum of all paths. Then I casted it to double before division by n*(n-1)/2. Got WA41. That happened because 'n' was 'int' and that multiplication overflowed. Then I casted 'n' to 'int64' inside that multiplication and got AC.