the segment you wrote to identify if a number is prime or not is much time killing,
rather you may use this function like this:
----------------------------------------------------------------------------
bool is_prime(int n)
{
    if(n < 2) return 0;
    else if(n == 2) return 1;
    else if(n%2 == 0) return 0;

    int root = sqrt(n);

    for(int i = 3; i <= root; i += 2) {
        if(n % i == 0) return 0;
    }

    return 1;
}
----------------------------------------------------------------------------
Hints: 2 is the only even number which is prime.

then you may store prime numbers in an array.
finally the program seems to be like this... :
---------------------------------------------------------------------------
#include <iostream>
#include <cmath>
using namespace std;

bool is_prime(int n)
{
    //if(n < 2) return 0;
    //else if(n == 2) return 1;
    //else if(n%2 == 0) return 0;    // this lines aren't necessary for this program.

    int root = sqrt(n);

    for(int i = 3; i <= root; i += 2) {
        if(n % i == 0) return 0;
    }

    return 1;
}

int main()
{
    int k, i, n, *num, prime_num[15001] = {0, 2};    // 2 is in 1st position.

    for(n = 3, i = 2; i <= 15000; n += 2)
        if(is_prime(n)) prime_num[i++] = n;

    cin >> k;

    num = new int [k];

    for(i = 0; i < k; i++) cin >> num[i];

    for(i = 0; i < k; i++)
        cout << prime_num[num[i]] << '\n';

    delete [] num;

    return 0;
}
---------------------------------------------------------------------------------

well done, bro.... :)
here's another way... >>
--------------------------------------------------------------------------------------
#include <iostream>
using namespace std;

int arr[100000];

int find_max(int n)
{
    int maximum = 0;

    for(int i = 0; i <= n; i++)
        if(maximum < arr[i]) maximum = arr[i];

    return maximum;
}

int main()
{
    int n[11], i, idx;

    arr[0] = 0;
    arr[1] = 1;

    for(idx = 2; idx < 100000; idx++) {
        if(idx & 1) {       // when idx is odd.
            i = (idx-1) / 2;
            arr[idx] = arr[i] + arr[i+1];
        }
        else {              // when idx is even.
            i = idx / 2;
            arr[idx] = arr[i];
        }
    }

    i = 0;
    while(cin >> n[i], n[i++] != 0);

    i--;
    idx = i;

    for(i = 0; i < idx; i++) {
        cout << find_max(n[i]) << '\n';
    }

    return 0;
}
-------------------------------------------------------------------------------------

Edited by author 07.06.2015 08:42

Use integral type (int, long, long long) or set precision in output format string printf("%0.0lf",a+b);

just use the integer

Well, maybe because you are checking it only for 4 Long values, It's working only for 4 exact values, not less, not more.


Edited by author 05.06.2015 13:04

Make sure s1 and s2 are within the limits of the problem.

I don't use any graphs algorithms,but only DFS.
If you try to solve this problem like me do:
1) break(remove) all cycles
2)Store elements of graph,witch belong to different component's of graph.
3) Connect one of them with others. (one - to every).
Not every to every!!!

Test
6
1000000 1000000
010000
100000
000100
001000
000001
000010



Edited by author 16.10.2009 21:29

Please explain, why my cod has WT7 and your not)

var a:array[1..1000] of integer;
i,n,k:integer;
Begin
 readln(n);
 for i:=1 to n do begin read(a[i]); end;
 k:=1;
 i:=2;
 while i<=n+2 do
 begin
 if a[i]=a[i-1] then
   begin
    i:=i+1;
    k:=k+1;
   end
 else
   begin
    write(k,' ',a[i-1],' ');
    i:=i+1;
    k:=1;
   end;
 end;
End.

Edited by author 26.11.2012 23:44

Edited by author 26.11.2012 23:44

or u can use
cout<<setprecision(2)<<fixed<<;

хотя нет, в три минуты тоже можно пожарить и разрезать ничего не надо

Edited by author 13.05.2014 01:30

Попробуйте
2 2 2 2 3 3 1
Ответ:
0.41
0.41

PLEASE NOTE what N and K represent;N represents colums and k represents rows.
My code was wrong because of this;Hope you can get help.