Совсем не могу понять почему не проходит 6 тест, проблема у меня не 59,9 как было у другого задовавшего вопрос...

#include <stdio.h>
#define c 65535
int main()
{
    long long k = 0;
    long long a[c];
    long n = 0,i,ot[c],j,t;

    a[0]=1;
    for (i=1; i<=c; i++)
    {
        a[i]=a[i-1]+i;
    }
    scanf("%ld",&n);
    for (i=0; i<n; i++)
    {
        scanf("%lld",&k);
        for (j=0; j<c; j++)
        {
            t=0;
            if (a[j]==k)
            {
                t=1;
                break;
            }
        }

        ot[i]=t;

    }
    for (i=0; i<n; i++)
    {
        printf("%ld ",ot[i]);
    }
    return 0;

}

That's how I get AC after so much time.

Edited by author 06.09.2013 22:11

For solving this problem I used advanced bipartite graph theory. But it work slowly(about 0.6sec). Very interesting how solve more quickly. Please, give me some hints.(My e-mail: scarlet.flower@list.ru)

On my VS 2012 C# compiler on test
1000 1000 1000
my program's output is
0.0000143861 288.6751345948
which doesn't seem to fit into precision (from the solution correctness criteria). But AC on timus under VS 2010 C# compiler.

Just interesting: there is no such test, timus VS 2010 is better than my VS 2012, or checker checks not equality of both numbers in output and answer, but some more weird thing?

What is test #19?

I think I found counter example for this, ie. set of points of which one is inside the 3D convex hull, but it's appropriate contenstant can win. If we are given for contestant with following speeds:

1 1 2
1 1 4
100 1 3
1 100 3
20 20 3

If we choose paths of length 100, 100, 3, than last contestant (20, 20, 3) can win with time 100/20 + 100/20 + 3/3 = 11. But, for this points 3D convex hull consists of (1, 1, 2), (1, 1, 4), (100, 1, 3) and (1, 100, 3).

Does anyone know if I'm wrong and why?

why wa #7
please help(

#include <stdio.h>

int main(void)
{
    int s;
    int p;
    float zo;

    int t = 0;

    scanf("%f",&zo);
    scanf("%d",&s);
    scanf("%d",&p);

    while(zo>s){

        zo -= zo*p/100;
        ++t;

    }

    printf("%d",t);


    return 0;
}

It's the first test where more than one memory cell needs to be modified.

Do I miss something in the problem statement?

E.g. is this test correct?

azkazkazkazkazkazkazoazv

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
+++++++++++++++++.+++++++++++++++++++++++++.<+++++++++++++++++++++++++++++++++++
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++.>>+++++
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
++++++++++++.<.<.>>.<.<.>>.<.<.>>.<.<.>>.<.<.>>.<.<++++.>>.<.----.

Observe than A[i] must be equal 0 or 1 ( no double ).
Just compute dp[v]. dp[v] is cost of traffic of subTree with root v if A[v] = 1.
If A[v]=0 than cost(of traffic of subTree with root v) equals zero.

Edited by author 03.09.2013 01:58

Edited by author 03.09.2013 01:59

Test:
2 1
2

answer: 1
Maybe your answer is 2? =)

Edited by author 02.09.2013 03:44

Edited by author 02.09.2013 03:46

My solution failed in test 30. How I can find what is wrong? Help me please

Edited by author 01.09.2013 23:49

Give me please 3 (third) test. I've tested with other tests from this forum - all works fine, but I have WA#3 problem...

I will show you only two combinations("x" means stone, that was
killed at last lead).
1). Those you can kill three stones
    oo   xo   oo   ox
    oo  oo   ox    o
    oo  oo   o     o

2). Those you can kill six stones
              o     oo    ox    o
    oooo   ooox   oox    oo    oo   ox   o
    oooo   ooo    oo     oo     xo   oo  ox

This are all you need to get AC. (But I cannot prove that
 if (M Mod 3 = 0) or (N Mod 3 = 0) answer is 2, I can only show how
make it)
P.S. Sorry for bad English.

#include<stdio.h>
#include<algorithm>
#include<list>
using namespace std;

int main(){
   char str[200000];
   cin>>str;
   bool litter[26];
   for(int i=0;i<26;i++)
   {
    litter[i]=false;
   }
   for(int i=0;i<strlen(str);i++)
   {
    litter[str[i]-'a']=!litter[str[i]-'a'];
   }
   list<char> clist;
   for(int i=0;i<strlen(str);i++)
   {
    if(litter[str[i]-'a'])
    {
    litter[str[i]-'a']=false;
    clist.push_back(str[i]);
    }
   }
   copy(clist.begin(),clist.end(),ostream_iterator<char>(cout));
   return 0;
 }
it do works on my own machine, How?

In some way,I got test1.
In the about 570th line,there is a mark"'".I don't know what is it for.I think it is by mistake.
It is same with some large test points.

Edited by author 05.04.2009 22:14

#include<iostream>
#include<stdio.h>
#include<math.h>
using namespace std;
struct point { double x;
               double y;
               double z; };
double pi=3.14159265358979323846;
int main () {
    point a,b,c;
    double tanga,tangb,distac,distbc,ang,arc,r,angpt;
    cin >> a.x >> a.y >> a.z >> b.x >> b.y >> b.z >> c.x >> c.y >> c.z >> r ;
    distac=sqrt((a.x-c.x)*(a.x-c.x)+(a.y-c.y)*(a.y-c.y)+(a.z-c.z)*(a.z-c.z));
    distbc=sqrt((b.x-c.x)*(b.x-c.x)+(b.y-c.y)*(b.y-c.y)+(b.z-c.z)*(b.z-c.z));
    tanga=sqrt(distac*distac-r*r);
    tangb=sqrt(distbc*distbc-r*r);
    angpt=acos(((a.x-c.x)*(b.x-c.x)+(a.y-c.y)*(b.y-c.y)+(a.z-c.z)*(b.z-c.z))/
                (sqrt((a.x-c.x)*(a.x-c.x)+(a.y-c.y)*(a.y-c.y)+(a.z-c.z)*(a.z-c.z))*
                 sqrt((b.x-c.x)*(b.x-c.x)+(b.y-c.y)*(b.y-c.y)+(b.z-c.z)*(b.z-c.z))))*(180/pi); //angle between the two vectors AC and BC
    ang=angpt-(((atan2(tanga,distac)+atan2(tangb,distbc))*(180/pi)));
    ang/=360;
    arc=ang*2*pi*r;
    printf("%.2lf",tanga+tangb+arc);
    cout << endl ;
    return 0;
}

I know that I have to check if there could be a straight line but I think that the sample test should be right for my program.
My program writes 19.78 not 19.71. Thanks in advance.

You need use function
one cycle,one function and Accepted

30000 30000
0    |
0    |
.    | => 30000 lines with '0' on each
.    |
0    |

There are solutions that may get Time Limit.

как пройти 45-ый тест