Please help.
You could not will lay out an approximate kind 8 tests.

Edited by author 12.05.2010 03:28

Sample? please

example N = 11, M = 3;

123
456
789
ab

the cell "3" can reach from the cell "9",
pushing an arrow key only once?
I mean "arrow down"?

This test looks something like
1
-1
______
-1

Good luck!

It is sooo easy :-)

Usual generic preflow algo got AC in 0.093 sec. Maybe, it worth to decrease timelimit?

if you are get wrong answer #1 or time limit #1 , probably your bug is that,you don't read input to end..I was get time limit #1..Because I didn't read input to end..when I find correct answer I break..then I found my bug and I read input to end..finally I got AC..If you don't read input to end , probably , your bug is that..you must read input to end..


Edited by author 17.06.2011 05:11

Pls, give me test 3.
I tested my prog with maybe 30 tests and every I passed.
But I have wa 3.
Why? Help me. Give me test 3 or some tests.


10x ;)
Sorry for poor english ;)

Why Crash (acces violant)
I use vector < map <int,int> > ! Are the problem in it?

P.S Sorry for my bad English

#include <cstdio>
#include <vector>
#include <algorithm>
#include <map>
#include <iostream>
#include <cstdlib>

using namespace std;

struct neww{
 int x,n;
};

int abss(int x)
 {
  if (x>=0) return x;
  else return (-x);
 }

bool funcsort(neww a, neww b)
 {
  if (a.x < b.x) return true;
  else return false;
 }

int main()
{
 int n,i,j,x,maxn;
 neww a[2200];
 scanf("%d",&n);
 vector < map <int,int> > b(12000);
 vector < map <int,int> > pi(12000);
 vector < map <int,int> > pj(12000);

 for (i = 0; i < n; ++i)
     {
      scanf("%d",&a[i].x);
      a[i].n = i + 1;
     }
 sort(a,a+n,funcsort);
 if (n == 1)
  {
   printf("1\n");
   printf("%d\n",1);
   exit(0);
  }

 for (i = n - 1; i >= 0; --i)
   {
      for (j = n - 1; j > i; --j)
         if (a[j].x-a[i].x > 0)
           {
            b[i][a[j].x-a[i].x] = max(b[i][a[j].x-a[i].x],b[j][a[j].x-a[i].x] + 1);
            if (b[i][a[j].x - a[i].x] = b[j][a[j].x-a[i].x] + 1)
             {
              pi[i][a[j].x-a[i].x] = j;
              pj[i][a[j].x-a[i].x] = a[j].x - a[i].x;
             }
           }
   }
 maxn = 0;
 int numb,ost;
 for (i = 0; i < n; ++i)
     for (j = i; j < n; ++j)
            if (b[i][abss(a[i].x - a[j].x)] > maxn)
             {
              maxn = b[i][abss(a[i].x-a[j].x)];
              numb = i;
              ost = abss(a[i].x - a[j].x);
             }
 printf("%d\n",maxn+1);
 printf("%d",a[numb].n);
 while (maxn)
   {
      int nm = pi[numb][ost];
      int ot = pj[numb][ost];

      printf(" %d",a[nm].n);

      numb = nm; ost = ot;
      --maxn;
   }
  printf("\n");

}

Edited by author 17.06.2011 01:42

Finally I`ve made it! It works properly,but when I submit it-Compilation error ...


#include<iostream>
#include<math.h>
using namespace std;

int main ()
{
    int p;
    cin>>p;
     for(int i=1;i<=sqrt(p);i++)
     {if(p%i==0 && p/i!=p) {cout<<i<<p/i<<" ";}
     if(p%i==1) return -1;}
     system("pause");
}

It's my solution:

#include<iostream>
#include<stdio.h>
#include<time.h>
using namespace std;
int main()
{
    #ifndef ONLINE_JUDGE
    time_t start,end;
    start=clock();
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    #endif
    int a,b,c,d=0,i,j;
    int *m,*m1;
    m=new int [100000000];
    m1=new int [100000000];
    cin>>a;
    for(i=0;i<a;i++)
        cin>>m[i];
    cin>>b;
    for(i=0;i<b;i++)
        cin>>m1[i];
    for(i=0;i<a;i++)
    {
        c=m[i];
        for(j=0;j<b;j++)
            if(m[i]==m[j]){
                m[j]=0;
                d++;
            }
    }
    cout<<d;
    delete [] m;
    delete [] m1;
    return 0;
}
Это решение даёт правильный ответ,но не проходит... Почему????

I don't how to slove this problem.who can help me?

i tryed all possible tests but i am still getting mistake on the fifth test. can somebody give me some tricky tests to help me?

What is test #10?

Now I have:
Input: 10 10 Answer: 10
Input: 13 14.1 Answer: 15

but:
float p, q;
scanf ("%f %f", &p, &q);
p/=100;
q/=100;

result: WA6

but:
float p, q;
scanf ("%f %f", &p, &q);
p=(float)((int)(p*100))/10000.00;
q=(float)((int)(q*100))/10000.00;
//Only 2 digits after dot

result: WA3

Why?

PS (ceil(p*cnt)-((int)(q*cnt)))>1e-7

Try to check situation, when rectangles touch and when rectangles don't have intersection.

var a:real;

begin

read(a);
writeln(sqrt(a):0:4)

end.

why is my answer wrong?? :(