Если на паскале в конце выводимой программы поставить "readln;", получите WA1

it says than "the sum of the first three digits....is equal with the sum of nthe last three".
In that case 45 first there 4+5+(null)=9
                last three 5+4+(null)=9
=>45 is a lucky number!!!
that means than there are (at nr 2) 11,12,13,...,99(100-1-10)=89 lucky numbers...
Can anyone explain me why they say yhere are only 10 numbers at example 2??
THANX

I got WA #41, can anyone give me some test case ?
Thank you :)

Mind the tests
000000 and 999999
999999+1 and 000000-1 don't belong to the definition of ticket

I got WA #4 with such code:

#include <cstdio>
#include <iostream>
#include <cmath>

using namespace std;

int main()
{
    int n;
    double a, sum = 0.0;

    scanf("%d", &n);
    for (int i = 0; i < n; i++)
    {
        scanf("%le", &a);
        sum += a;
    }

    printf("%.15e", sum / n);

    return 0;
}

WA #5 with such code:

#include <cstdio>
#include <iostream>
#include <cmath>

using namespace std;

int main()
{
    int n;
    double a, sum = 0.0;

    scanf("%d", &n);
    for (int i = 0; i < n; i++)
    {
        scanf("%le", &a);
        sum += a;
    }

    printf("%.7e", sum / n);

    return 0;
}

WA #6 with such code:

#include <cstdio>
#include <iostream>
#include <cmath>

using namespace std;

int main()
{
    int n;
    double a, sum = 0.0;

    scanf("%d", &n);
    for (int i = 0; i < n; i++)
    {
        scanf("%le", &a);
        sum += a;
    }

    printf("%.8e", sum / n);

    return 0;
}

AC with such code:

#include <cstdio>
#include <iostream>
#include <cmath>

using namespace std;

int main()
{
    int n;
    double a, sum = 0.0;

    scanf("%d", &n);
    for (int i = 0; i < n; i++)
    {
        scanf("%le", &a);
        sum += a;
    }

    printf("%.15e", sum / n);

    return 0;
}

But, connecting to the problem statements, "The offset should be printed with at least six digits after decimal point."

I'm using pure Monte-Carlo method.

Here is my code (C#):

[Code deleted by author]

I've found my bug...

P.S. For the pure Monte-Carlo method it is enough 10^5 points and Double type.

Edited by author 13.06.2010 19:25

How to show 6 digits after the point using double/float variable?

You just need to check all possible numbers of L (1..10000) to solve this. So you get O(n^2) solution.

Standard template library is not available or what?
i am getting compilation error for this problem.This is my code
<code>
#include<iostream>
#include<algorithm>
using namespace std;
class node{
        public:
                int id,prob;
                bool operator<(const node &n)const{
                        return (prob>n.prob);
                }
};
int main()
{
        int n;
        scanf("%d",&n);
        node arr[n];
        for(int i=0;i<n;i++){
                scanf("%d%d",&arr[i].id,&arr[i].prob);
        }
        sort(arr,arr+n);
        for(int i=0;i<n;i++)
        {
                printf("%d %d\n",arr[i].id,arr[i].prob);
        }
        return 0;
}
</code>

Edited by author 11.06.2008 17:19

Tell me please!
Is it greedy???

Can anybody give me some good test?
I can't find any bug

Admins,i think i need you help

I'm sure I got WA because of reading string.
Please help me, thanks.
Here is my code to read string :
...
fscanf(stdin,"%d\n",&m) ;
   for(int i=0;i<m;i++) {
       char s[1000] ;
         int j = 0 , k = 0 ;
         fscanf(stdin,"%s",&s) ;
         while ( s[j] != ',' ) a[i].x = a[i].x * 10 + s[j++] - 48 ;
         j++ ;
         while ( s[j] != ':' ) a[i].y = a[i].y * 10 + s[j++] - 48 ;
         j++ ;
         while (j < strlen(s) ) {
               while ( s[j] != '-' ) b[i][k] = b[i][k] * 10 + s[j++] - 48 ;
               j++ ;
               while ( j < strlen(s) && s[j] != ',' ) c[i][k] = c[i][k] * 10 + s[j++] - 48 ;
               k++ ;
               if ( s[j] == '\n' ) { j++ ; break ; }
               j++ ;
         }
         deg[i] = k ;
     }

The obvious solution is too simple.
You should ask for the smallest numbers. That solution is also easy enough, so it shouldn't be a problem for most people.

Anyway, maybe the purpose of this problem is to teach people to use __int64. In that case simpler is better :)

Edited by author 22.02.2009 21:50

[code deleted]

Edited by author 20.10.2006 23:12

Edited by moderator 01.12.2019 21:13

Can anybody give me some correct input-output data for testing?

I solve this problem about half of year.
I use Binary Heap to store n/2+2 elemets at first.
If you have WA 6-8 try to check your Binary Heap in some random tests with odd and even N (10 or 11)
My wrong Binary Heap work different in other order of elements :)
For example
10
1 2 3 4 5 6 7 8 9 10

10
10 9 8 7 6 5 4 3 2 1

10
1 6 8 3 9 5 2 10 4 7

Answers are 5.5

11
1 2 3 4 5 6 7 8 9 10 11

11
11 10 9 8 7 6 5 4 3 2 1

11
11 10 3 8 9 7 6 2 1 5 4

Answers are 6.0

P.S: Escape integer overflow :)

Edited by author 09.10.2009 02:56

I passed for G (N ^ 2). Is this normal? What is the asymptotic behavior of a normal solution?

Anybody, can tell me what the test 9?