OJ Board

Hint: The result is a wellknown constant

The hexadecimal expansion of 1/11 isnt 0.0F0F.. that is the hexadecimal expansion of 1/17 ...

You dont have to use K separations, the 2 ways are:

(1+1) (10) with 1 separation
(1)(1)(10) with 2 separations

My Guess is some input like:

Try to read m and n with cin >> m >> n and surprisingly u will get wa.

Does anyone know whats the difference between cin and scanf("%d.. in reading integers?

There is definitly something wrong around some inputs for m and n which scanf can handle somehow and cin not.

Dont know whats wrong with input yet. But this simple input checker gets tle :-) #include <iostream> #include <string> using namespace std; string s; char c; int t,d,q,m,n,i,j; int main() { cin >> t; while(t--) { cin >> d; while(d--) { cin >> s; } cin >> q; while(q--) { cin >> m >> n; cin.get(); whi...

That was allready known:

http://online-judge.uva.es/board/viewtopic.php?t=9141

WORD w[100000];

why is that array only 100000 big ?

if all words differ in first 3 letters and are 100 letters long u need 98*words entrys

i'll write and submit an input checker tomorrow

int d[2000*d_rate*l_rate], l[2000*l_rate*d_rate];

that arrays are to big.
if d_rate=l_rate=100 they are 76 MB

an array is enough

first solver under 0.1 s with scanf ^^

It reads a string from input into the chararray with ending '\0'
i always use an array long enough and there were no bugs yet

Thnx mjf ... 0.150 seconds after that without using optimized io :-) Official solution uses a non-logical algorithm, that says that you can't go from one island to another if they overlap. We're going to change the input, so that no 2 islands intersect. And a negative island number is logical? ... 0...

My Solution takes about 40 sec with same answer

The sequence is composed by integers, each integer is greater than or equal to 1 and less than or equal to 1000.

Wrong ... there are inputs that are not in that range ... another way making an easy problem hard ...

The correct answer is 81. Lets take this simple Programm to show you the idea: IF ELSE END_IF IF ELSE END_IF IF ELSE END_IF ENDPROGRAM Here all IF-Conditions can be true, short: TTT Or the first two are true and the third false, short: TTF And so on to FFF So at all there are 2*2*2=8 possibilities. ...