#include<stdio.h>
#include<stdlib.h>


int main()
{
long unsigned int a[15000][2],i,j,n,temp,b,c;
scanf("%ld",&n);

for(i=0;i<n;i++)
{
scanf("%lu%lu",&b,&c);
if(b<=10000000&&c<=100)
{
a[i][0]=b;
a[i][1]=c;
}
else
{
return 0;
}
}


for(i=0;i<n;i++)
{
for(j=0;j<n-i;j++)
{
if(a[j][1]<a[j+1][1])
{
temp=a[j][0];
a[j][0]=a[j+1][0];
a[j+1][0]=temp;

temp=a[j][1];
a[j][1]=a[j+1][1];
a[j+1][1]=temp;
}
}
}
for(i=0;i<n;i++)
{
for(j=0;j<2;j++)
{
printf("%d ",a[i][j]);
}
printf("\n");
}

return 0;
}

Edited by author 27.03.2013 18:08

her's my cod! Please help!
using System;

namespace ConsoleApplication1
{
    class Program1563
    {
        static void Main()
        {
            int n = int.Parse(Console.ReadLine());
            string str;
            string[] strl = new string[n];
            int s = -1;
            for (int i = 0; i < n; i++)
            {
                str = Console.ReadLine();
                strl[i] = str;
                Console.WriteLine("i="+i);
                for (int j = 0; j < i; j++)
                {
                    if (str == strl[j])
                        s++;
                }
            }

            Console.WriteLine(s);
                    Console.ReadLine();
        }
    }
}

When you compare in the sort, try using epsilon = 1e-8.

When I used 1e-13, I keep getting WA#3.

I use O(n^3), but i meet time  problem. Can you help me?

#include <iostream>

int ClickCount(char c);

using namespace std;

int main()
{
    char Buf[1001];
    cin.getline(Buf,1001);

    int Result=0;
    int i=0;
    while (Buf[i]!='\0')
    {
        Result+=ClickCount(Buf[i]);
        i++;
    }

    cout << Result;

    //system("pause");

    return 0;
}

int ClickCount(char c)
{
    char AlphaBet[]="abcdefghijklmnopqrstuvwxyz*.,!# *";
    for (int i=0;i<=sizeof AlphaBet;i++)
    {
        if (c==AlphaBet[i]) return i%3+1;
    }
    return 0;
}

It took me 2 minutes to finish it.

if N=3 k=10
is 011 valid(i mean may the valid number start with 0?)

import java.util.*;
public class LavinInteractive {


    public static void main(String[] args) {

        int n;
            Scanner rc = new Scanner(System.in);
            System.out.print("Введите число: ");
            n = rc.nextInt();

            if (( n >= 1 ) && (n <= 4)) {
                System.out.print("few");
            }
            else if (( n >=5 ) && (n <= 9)) {
                System.out.print("several");
            }
            else if (( n >= 10 ) && (n <= 19)) {
                System.out.print("pack");
            }
            else if (( n >= 20 ) && (n <= 49)) {
                System.out.print("lots");
            }
            else if (( n >= 50 ) && (n <= 99)) {
                System.out.print("horde");
            }
            else if (( n >= 100 ) && (n <= 249)) {
                System.out.print("throng");
            }
            else if (( n >= 250 ) && (n <= 499)) {
                System.out.print("swarm");
            }
            else if (( n >= 500 ) && (n <= 999)) {
                System.out.print("zounds");
            }
            else if ( n >= 1000 ) {
                System.out.print("legion");
            }



    }

}

могут ли сигареты накладываться друг на друга?
например:

0 0 0 2
0 1 0 3

po4emu u menia Time Limit Exceeded? (

Edited by author 12.06.2012 08:19
#include <iostream>
using namespace std;
int main()
{
 int a[4000],a2[4000],a3[4000],i,j,l,n,n2,n3,k=0;
 cin>>n;
 for(i=1;i<=n;i++)
 cin>>a[i];
                cin>>n2;
                          for(j=1;j<=n2;j++)
                               cin>>a2[j];
cin>>n3;
for(l=1;l<=n3;l++)
cin>>a3[l];
          for(i=1;i<=n;i++)
          {
              for(j=1;j<=n2;j++)
              {
                  for(l=1;l<=n3;l++)
                  {
                      if(a[i]==a2[j] && a[i]==a3[l]) k++;
                  }
              }
          }
cout<<k;
return 0;
}


Edited by author 12.06.2012 08:25

My program(LANG c)is in the following:
#include<stdio.h>
#include<stdlib.h>
#include<cmath>
int k[60000],value,i,n;
int main()
{
    scanf("%d",&n);
    for(i=1;i<=n;i++)
       scanf("%d",&k[i]);
    for(i=1;i<=n;i++)
    {
         if(k[i]==0) printf("0 ");
         else
         {
              value=(int)sqrt((k[i]-1)*2);
              if((value*(value+1))==2*(k[i]-1))
                 if(i==n)  printf("1\n");
                 else printf("1 ");
              else
                 if(i==n) printf("0\n");
                 else printf("0 ");
              }
    }
    return 0;
}

BUT it can't compile,said that
"
d9b0f699-b537-45eb-b286-ce7765096c87d9b0f699-b537-45eb-b286-ce7765096c87(16) : error C2668: 'sqrt' : ambiguous call to overloaded functionS:\checker\compile\vc10\include\math.h(589): could be 'long double sqrt(long double)' S:\checker\compile\vc10\include\math.h(541): or 'float sqrt(float)' S:\checker\compile\vc10\include\math.h(127): or 'double sqrt(double)' while trying to match the argument list '(int)'
"

BUT in my computer(Windows 7),I can compile and run it,and the answer is right.
So I don't know where is wrong?And I don't know how to modify it.Help me,please!

Edited by author 22.02.2013 15:09

There is some trash in the end of file.

hello,
I wrote the code for this problem in python. I sure my logic is right, but do not know why i get a wrong answer. I have pasted my code below. If someone could help me find where the bug is .. it would be of great use

#timus 1005
import sys,math
class Timus1005:
    def __init__(self):
        pass

    def minimumDiffrence(self,weights):
        weights.sort()
        if (len(weights) == 1):
            return weights[0]

        if (len(weights) == 2):
            return (weights[1] - weights[0])

        weights.reverse()

        pile1 = 0
        pile2 = 0

        for stone in weights:
            added = False
            if (pile1 < pile2) and added == False:
                pile1 = pile1 + stone
                added = True

            if (pile1 > pile2) and added == False:
                pile2 = pile2 + stone
                added = True

            if (pile1 == pile2) and added == False:
                pile1 = pile1 + stone
                added = True

        return int(math.fabs(pile1 - pile2))

if __name__ == "__main__":
    w = []
    for line in sys.stdin:
        for word in line.split(' '):
            w.append(int(word))

    p = Timus1005()
    print p.minimumDiffrence(w[1:])

My greedy algorithm have 'accepted', but gives wrong answer for this test:

8
a b c
d e f
g i k
b l m
l m i
i k c
d e c
e f k

The number of ways to tile an MxN rectangle with 1x2 dominos is
2^(M*N/2) times the product of
{ cos^2(m*pi/(M+1)) + cos^2(n*pi/(N+1)) } ^ (1/4)
over all m,n in the range 0<m<M+1, 0<n<N+1.

I have found this formula in internet and I would like to
know is there a way to use it for solving this problem?

My first solution printed 2, if first player wins and the number of rocks is even and printed 1 if it's odd. And this solution got WA9. After printing "some variable" mod 3 the solution got AC. But why did my first idea fail?

I realize my idea on Java and Pascal.
On java i got TLe on 22 test, but on Pascal i got AC with time 0.069.
Not good when problem depends on the language.

It's not me posted if you can remove it? Thanks!

Edited by author 24.03.2013 19:47

I test my code at my computer,I got a test which is filled by period string "abcde" and its length is 4000,and my program run only 0.25s.
My computer is:
Memory:3GB
System:Windows XP
CPU:1.66GHZ
But I got tle at #32...Why???
Here is my code.
var
        i,j,xx,ox,t,l:longint;
        s:ansistring;
        f,x:array [1..4010] of longint;
        o:array [0..4010] of boolean;
function pdhw(l,r:longint):boolean;
var
        i:longint;
begin
        if l=r then exit(true);
        for i:=l to (l+r)>>1+1 do
          if s[i]<>s[r+l-i] then exit(false);
        exit(true);
end;
begin
        readln(s);
        l:=length(s);
        for i:=1 to l do
        begin
          f[i]:=maxlongint;
          if pdhw(1,i) then f[i]:=0
          else for j:=1 to i-1 do
                 if (f[i]>f[j]+1)and(pdhw(j+1,i)) then begin
                   f[i]:=f[j]+1;
                   x[i]:=j;
                 end;
        end;
        writeln(f[l]+1);
        xx:=l;
        repeat
          ox:=xx;
          xx:=x[xx];
          if ox=l then o[xx+1]:=true
          else o[xx+1]:=true;
        until xx=0;
        o[1]:=false;
        for i:=1 to length(s) do
          if o[i] then write(' ',s[i])
          else write(s[i]);
end.