Try a to test with two same masses:

5
11
13
21
10000
10000

Then when you will understand what was your problem you will use

std::multiset<double, std::greater<double>> s;

Cheeres! ;)

you don't need to use BigInteger =)

Thanks, your hint really helped me. The same thing in Java. I had TLEon15. After I replaced Scanner with my custom input parsing I had got AC.

O(M + N) is optimal. But maybe you can succed with your asymptotic if use fast I/O

I also can't understood why Java is disallowed now. Even if java solution is not so fast as C/C++/Pascal, it's iteresting to make some optimisations to java solution, to get accepted.

Please, accept Java for this problem.
Now empty Java 1.7 program takes only 116Kb, so 750kb would be enought to make a solution.

Edited by author 05.07.2013 13:04

There is one more case when N>2 and all languages are the same.

he he...
finaly i passed this test
you should be carefull with 32000
its better to use 32002 :)
Good Luck!

Yes. We can decrease flow on some pipeline( if the flow on some pipeline is greater zero ). And this operation is free!!! =)

my code:


#include <string>
#include <iostream>
using namespace std;
int main()
{
    string line;

    getline(cin, line);

    static bool sing = false;
    for(int i = 0; i < line.length(); i++)
    {
        if(sing)
        {
            if((line[i] >= 'A') && (line[i] <= 'Z'))
            {
                char temp = line[i] - ('A'-'a');
                cout << temp;
            }
            else if((line[i] == '.') || (line[i] == '!') || (line[i] == '?'))
            {
                cout << line[i];
                sing = false;
            }
            else if(line[i] == 10 || line[i] == 26)
            {
                cout << '\n';
            }
            else
            {
                cout << line[i];
            }
        }
        else
        {
            if(line[i] >= 'A' && line[i] <= 'Z')
            {
                cout << line[i];
                sing = true;
            }
            else if(line[i] == 10 || line[i] == 26)
            {
                cout << '\n';
            }
            else if((line[i] == '.') || (line[i] == '!') || (line[i] == '?'))
            {
                cout << line[i];
            }
            else
            {
                cout << line[i];
            }
        }
    }
    return 0;
}

Once you've reduced it that far, you can use algebra to show that all the procedure in the for loop really does is swap the variables. That coupled with the conditions of the for loop give you the O(1) solution.

Edited by author 01.07.2013 01:03

me too
help