Does the problem statement meant that each member should have all his friends in the other team OR each member should have at least 1 friend in the other team ?
If the first statement is correct the sample output does not hold.
Please clarify.
Thanks in advance

Edited by author 17.02.2013 01:32

#include <iostream>
#include <cmath>
int i=0;
using namespace std;
void f();
int main()
{

    cout.setf( ios::fixed );
    cout.precision( 4 );
    f();
    return 0;
}
void f()
{
    long int a;
    if (cin >> a) f();
    if (i!=0) {
    cout << sqrt((double)a) << endl;}
    else i++;
}

use long long int instead of int , it may help..

Edited by author 20.08.2006 07:47

This is a problem with possible loops in decision. That means your decision will be based on others' decision, which might  again be based on yourself. I think the best way to solve this sort of problems is to use ITERATION.

I believe that some people could solve this directly. If you are like this, please post your idea or formula, I want to know it very much.

During iteration, we must pay attention to the initialization value. If you have a bad init-value, you will not be able to  move even one step in this problem. In my opinion, the init-value could be set as: when someone is starting a duel, then he could win this duel with prob 1.

By the way, iteration will stop very quickly, as the calculation is unusual: there is function "MAX". So do not worry about the complexity of the problem.

At last, some test-data here:

1.0 0.8 0.5
0.30000000 0.17777778 0.52222222

1.0 0.5 0.0
0.75000000 0.25000000 0.00000000

1.0 1.0 0.0
0.50000000 0.50000000 0.00000000

1 0.9 0.8
0.11000000 0.08265306 0.80734694

1 0.99999 0.99998
0.00001000 0.00001000 0.99998000
(Suprising or not?)

1 0.0001 0.0002
0.99975002 0.00014998 0.00010000

So, Good luck.

when you calculate q you must use (res-(1e-7))*q/100.0;

is in your example abababa right answer aba or only a???
because a is 4 and aba is 3.

"choosing" -> "chosen"
"divided" -> "divisible"
"satisfied" -> "satisfying the"
"is really completely defines" -> "completely defines"

These aren't just typos, they impede the understanding.

I cannot understand the problem. The question is phrased pOorly !

If there is anyone, who can tell me, the question clearly, I am ready to phrase it in a better way.

The "YES" come from two ways:1) when there is not have answer.txt;2) when answer.txt contain it.but i can't figure out this related to pass test! pass test should be the output.txt's content is same to  answer.txt. how to explain the hint's explanation?

No matter when I used double or long double, my code still got TLE.

Opt[Now][j][0] = Opt[Pre][j][0] + Opt[Pre][j][1] + Opt[Pre][j][3];
Opt[Now][j][1] = Opt[Pre][j - 1][0] + Opt[Pre][j - 1][2] + Opt[Pre][j - 1][1];
Opt[Now][j][2] = Opt[Pre][j][2] + Opt[Pre][j][3];
Opt[Now][j][3] = Opt[Pre][j - 1][2] + Opt[Pre][j - 1][3];

I had to write a Pascal code using extended and got AC.
Could anyone tell me how to AC in C++?

I don't know why this solution is correct, but it got accepted.
I thought that this solution will get TL.
Author of problem: easy tests?

Who couldn't find solution try to understand this code:

#include<map>
#include<cmath>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<string.h>
//#include<windows.h>
#include<algorithm>
#define y1 abcde
#define sqr(x) ((x)*(x))
#define INF 20000000000000000
using namespace std;

   string s;
   int i,n,ans,j,l,r;

   bool check(int l,int r)
{
        for(int i = 0; i < (r - l + 1) / 2; i++)
        if (s[l+i] != s[r-i]) return false;
    return true;
}
int main()
{
   //srand(GetTickCount());
   #ifndef ONLINE_JUDGE
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
   #endif

    cin>>s;
    n = s.size(); i++; s = '#' + s;
    for(i = 1; i <= n; i++)
     for(j = i; j <= n; j++)
    {
             if (check(i,j))
              if (j - i + 1 > ans)
                   ans = j-i+1,
                   l = i,
                   r = j;
    }
     for(i = l; i <= r; i++) cout<<s[i];
    return 0;
}

Edited by author 13.02.2013 14:44

Edited by author 13.02.2013 14:45

3
1 1 1

answer
3 2

What is algorithm? What formulas, theorems, equations and hints must I use?

using System;
class Program
{
    static void Main()
    {
        int N = int.Parse(Console.ReadLine());
        if (N > 0 && N <= 100)
        {
            string[,] IData = new string[N, 3];
            string[] buf = new string[2];

            for (int i = 0; i < N; i++)
            {
                buf = Console.ReadLine().Split(' ');
                IData[i, 0] = buf[0];
                IData[i, 1] = buf[1];
                IData[i, 2] = Convert.ToString(i + 1);
            }

            String[] Win = new string[N];
            string[] Lider = new string[2];
            Lider = IData[0, 1].Split(':');
            int k = 0;

            for (int i = 1; i < N+1; i++)
            {
                buf = IData[i, 1].Split(':');
                if (int.Parse(Lider[0]) > int.Parse(buf[0]))
                {
                    if (i == 1 && (60 - double.Parse(buf[1].Replace(".", ",")) + double.Parse(Lider[1].Replace(".", ",")) < 30))
                    {
                        Win[k] = IData[0,0];
                        k++;
                    }
                    Win[k] = IData[i, 0];
                    k++;
                    Lider = IData[i, 1].Split(':');

                }
                if (int.Parse(Lider[0]) == int.Parse(buf[0]))
                {
                    if (double.Parse(Lider[1].Replace(".", ",")) > double.Parse(buf[1].Replace(".", ",")))
                    {
                        double tmp = double.Parse(Lider[1].Replace(".", ",")) - double.Parse(buf[1].Replace(".", ","));
                        if (i == 1 && tmp < 30)
                        {
                            Win[k] = IData[0, 0];
                            k++;
                        }
                        Win[k] = IData[i, 0];
                        k++;
                        Lider = IData[i, 1].Split(':');
                    }
                }
                if (int.Parse(Lider[0]) < int.Parse(buf[0]))
                {
                    Win[k] = IData[0, 0];
                    k++;
                }
            }
            Console.WriteLine(k);
            Array.Sort(Win);
            for (int i = 0; i < k; i++)
            {
                Console.WriteLine(Win[i]);
            }
        }
    }
}

Edited by author 12.02.2013 23:07

Несмотря на то, что задача просто решеется с помощью std::set, я написал сжатый бор. Первый раз писать немного сложно, но весело. А теперь ещё и просто.

а если в вершинах бора хранить не map<string, node*> а попроще map<pair<int, int>, node*>, то я думаю результат был бы более впечатлительным

P.S. решайте сжатым бором

Accepted  0.312     2 904 КБ

Edited by author 11.01.2012 15:06

my program output result:

1 1000000000 3 13   -> 84

1 1 2 2 ---> 0

1 100 2 4 ->6

1 300 4 8 ->0

1 400 4 2 ->111

1 2147483647 13 3 --> 77520

1 2147483647 20 2  --> 84672315

1 3 1 2 --> 2

is right?

if full right , but i alway  wa #7,   who can give much testdata to me ?


Edited by author 03.09.2008 10:39

Edited by author 05.09.2008 02:08

i've got AC but i omitted the case "impossible" (ie print "0"). it has always solution?

e.g. the input is (if I'm not wrong it's valid input):

3
Isenbaev A B
A A A
B B B

What is the correct answer?

#include <cstdio>
#include <deque>
#include <queue>
#include <vector>
#include <string>
#include <set>
#include <stack>
#include <algorithm>
#include <cmath>
#include <iostream>
#include <fstream>
#include <climits>
#include <cfloat>
#include <map>
#include <cstring>

#define FOR(i,s,e) for(long i = s; i < e; ++i)
#define FORD(i,e,s) for(long i =e-1; i>=s; --i)
#define read(a) scanf("%ld",&a);
#define readf(a) scanf("%lf",&a);
#define read2(a,b) scanf("%ld %ld",&a,&b)
#define read2f(a,b) scanf("%lf %lf",&a,&b)
#define write(a) printf("%ld ",a);
#define writef(a) printf("%lf ",a);
#define write2(a,b) printf("%ld %ld ",a,b)
#define write2f(a,b) printf("%lf %lf ",a,b)
#define writeln(a) printf("%ld\n",a);
#define writelnf(a) printf("%lf\n",a);
#define writeln2(a,b) printf("%ld %ld\n",a,b)
#define writeln2f(a,b) printf("%lf %lf\n",a,b)
#define nln printf("\n")
#define elif else if
#define MOD 1000007
#define EPS 1e-9
#define INF 2117117117
#define mp(a,b) make_pair(a,b)
#define dist(x1,y1,x2,y2) sqrt((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1))
#define pb(a) push_back(a)

using namespace std;

int main()
{
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    long n;
    read(n); scanf("\n");
    map<string, long> m;
    vector<long> g[318];
    long c=0;
    string s1,s2,s3;
    FOR(i,0,n)
    {
        getline(cin,s1,' ');
        if (m.find(s1)==m.end())
            m.insert(mp(s1,c++));
        getline(cin,s2,' ');
        if (m.find(s2)==m.end())
            m.insert(mp(s2,c++));
        getline(cin,s3,'\n');
        if (m.find(s3)==m.end())
            m.insert(mp(s3,c++));
        long a=m.find(s1)->second,b=m.find(s2)->second,c=m.find(s3)->second;
        g[a].pb(b); g[a].pb(c); g[b].pb(a); g[b].pb(c); g[c].pb(a); g[c].pb(b);
    }
    long d[318];
    FOR(i,0,c)
        d[i]=-1;
    long s = m.find("Isenbaev")->second;
    d[s]=0;
    queue<long> q;
    q.push(s);
    long v;
    while (!q.empty())
    {
        v=q.front();
        q.pop();
        FOR(i,0,g[v].size())
        {
            if (d[g[v][i]]==-1)
            {
                d[g[v][i]]=d[v]+1;
                q.push(g[v][i]);
            }
        }
    }
    map<string, long>::iterator it;
    for( it = m.begin(); it != m.end(); ++it)
    {
        printf("%s ",it->first.c_str());
        if (d[it->second]==-1)
            printf("undefined");
        else
            write(d[it->second]);
        nln;
    }
    return 0;
}

Why??