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вернуться в форумSeems 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 if
2 3 - imput data
1 common can
(2+3)-1=4 - amount of cans readln(n,m);
writeln(m-1,' ',n-1); 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 Yes it has helped me after 9 more year
Edited by author 03.07.2020 00:58 |
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