Seems an easy problem, but I can't understand it...

4 7 - input data
 1  - common can
(4+7)-1=10 - amount of cans

How can you get "1-common can"?

Or the problem says there must be one common can?

At some moment it happened so that they shot the same can - last can

Still a little puzzled... Anyway I have got AC, thx for your help :)

why I am wrong on test2,it seems very easy

"The 12th High School Pupils Collegiate Programming Contest of the Sverdlovsk Region (October 15, 2005)"

What a mockery  ?!?!

cin >> h >> l;
cout << l-1 << " " << h-1;

And that's all ... (0.001 sec and 220Kb memory used :) )

P.S. 137Kb if it's transformed for C

Edited by author 06.10.2006 00:10

readln(n,m);
writeln(m-1,' ',n-1);

if
2 3 - imput data
1 common can
(2+3)-1=4 - amount of cans

I can explain it easily.

1 1 1 1 "1" 1 1 1 1 1 1 1

the quoted 1 is the can which they will shot it at the same time.

hope i helped after 4 years :D

sdsdksk skdk skdkfore

haha thanks

Yes it has helped me after 9 more year

Edited by author 03.07.2020 00:58