Please look at the page:
http://tieba.baidu.com/f?kz=632211990

Please look at the page:
http://tieba.baidu.com/f?kz=632211990

Can anybody tell me what's wrong with this?
When I test the program it works fine.
I end the input using ctrl-D , Is that wrong?
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>

/*function to reverse a string*/
void strrev(char string[256])
{
 char reverse[256];
 int i,j;
 i=strlen(string);

 for (j=i-1;j>=0;j--)
  {
    reverse[i-1-j]=string[j];
  }
  reverse[i]='\0';
  strcpy(string,reverse);
}


int main(void)
{

 int k=0,i=0,j,different=0,start=0;
 int c;
 char stream[256];
 char num[256];
 char copy[256];
 double number;
 while((c=getchar())!=EOF)
  {
   stream[i]=c;
   j=i;
   i++;
  }

 for (k=j;k>=0;k--)
  {
   if(isdigit(stream[k])!=0)
     {
       if(isdigit(stream[k-1])!=0)
        {
          num[start]=stream[k];
          start++;
        }
       else
        {
          num[start]=stream[k];
          num[start+1]='\0';
//          printf("%s\n",num);
          strrev(num);
//          printf("%s\n",num);
          strcpy(copy,num);
          number=strtod(copy,NULL);
          printf("%0.4f\n",sqrt(number));
          start=0;

        }
     }
  }

  return 0;

}

What is Fail (validator)? Checker returned this error on test 4 for my some crazy solution.

Got real pleasure when was solving it. But why are limitations so small?

ignore it..just my mistake

Edited by author 22.08.2009 13:35

Is it a precision problem?
who can give me some useful tests?

Can you say me what answer does my program give at test #2;
 I know this test is 1 2 3 4 0 and at my computer i have answer 1 2 3 4 and I'm sure it's right answer; but I have WA2 at Timus, and I really don't like it.

You can submit this program to see why test 2 is 1 2 3 4 0:

#include <iostream>

using namespace std;

int main() {
    int a, b, c, d, e;
    scanf("%d",&a);
    if (a==3) {printf("2 2 3 4 4 5 3");return 0;}
    scanf("%d%d%d%d",&b,&c,&d,&e);
    if (a==1 && b==2 && c==3 && d==4 & e==0) while(1);
}

You'll get TL 2.

Edited by author 21.08.2009 23:12

i decompose graph into connected components (by one dfs)
after that i count diameter of each component(by one special_dfs) and refresh global_variable if need. my programm works at all my tests and
sample test too.

I dont understand the problem itselk. what does K stands for?

Hi, for the "F" parameter, I can see that it is the hight of the binary tree. But in my AC program, I didn't even use it. below is my program, anything I've missed?

======================
code deleted
==============================

Edited by moderator 01.10.2022 12:26

Я раз 10 прочитал задачу, и до сих пор не понял, что от меня требуется(((((

Автору зачОт))))

What is the trick???

I have WA3. May be I don't understand this problem.
#include <stdio.h>

int main()
{
    int n,a[20],i,s=0,rez=0;
    scanf("%i",&n);
    for (i=0; i<n; i++)
    {
        scanf("%i",&a[i]);
        s=s+a[i];
    }

    /*for (i=0; i<n; i++)
        printf("%i",a[i]);*/

    int maxs,maxi,j,sdop,k;
    bool ok=true;

    while (ok)
    {
        maxs=0;

    for (i=0; i<n; i++)
    {
        sdop=0;
        for (j=i; j<=i+2; j++)
        {
            if (j>n-1) sdop=sdop+a[j%n];
            else sdop=sdop+a[j];
        }
        if (sdop>maxs)
        {
            maxs=sdop;
            maxi=i;
        }
    }

    s=s-maxs;
    rez=rez+s;

    int d1,d2,d3;
    d1=maxi;
    if (maxi+1>n-1) d2=(maxi+1)%n;
    else d2=maxi+1;
    if (maxi+2>n-1) d3=(maxi+2)%n;
    else d3=maxi+2;

    a[d1]=0;
    a[d2]=0;
    a[d3]=0;

    ok=false;
    k=0;
    while (!ok && k<n)
    {
        if (a[k]!=0) ok=true;
        else k++;
    }

    }
    printf("%i",rez);

    return 0;
}

Check please test 33.
It's very strange that dp ( subset with given sum ) don't work.
Maybe there is a bad checker.

Can anybody explain me, where am I wrong?

Definitions:
ac, ab, bc - corresponding lengths
acb - angle between (ca) and (cb)

Such code:
...
acb = acos((ac*ac + bc*bc - ab*ab) / 2.0 / ac / bc);
..

gets AC.

Such code:
...
double spv(point x1, point x2, point y1, point y2){
    int i;
    double res = 0;
    for (i=1; i<=n; i++){
        res += (x2.coord[i] - x1.coord[i]) * (y2.coord[i] - y1.coord[i]);
    }
    return res;
}
...
acb = acos(spv(c,a,c,b)/ac/bc);...
...

gets WA25.

I had always thought that these two fragments must yield the same result. Why don't they?

I use n^4 algorythm and works for 0.015 secs. :P

please, any hint