5 rows of code to solve this

My Endlish is not that good, could anyone translate it into Russian, pls?

Isn't it?

заходит решение
берем пары соседних чисел и выводим минимум из их нодов(

#include <iostream>
#include <vector>

using namespace std;

int gcd(int a, int b)
{
    if(a == b)
        return a;
    while( a != 0 && b !=0)
        if(a > b)
            a%=b;
        else b%=a;
    return a+b;
}
int main()
{
    int N;
    cin >> N;
    int array[1000];
    for(int i = 0; i < N; i++)
        cin >> array[i];
    if(N == 1) cout << array[0];
    else{
    vector <int> mins;
    for(int i = 0; i < N-1; i++)
        mins.push_back(gcd(array[i], array[i+1]));
    int min = mins[0];
    for(int i = 0 ;i < mins.size(); i++)
        if(min > mins[i])
            min = mins[i];
    cout << min;
    }
    return 0;
}
это вроде не камильфо!

Confusing question: HOW ARE THE BLANKS CHOSEN?

Let's see example
2 3 4 - initial blanks
I choose blanks 2 and 4 - they are different so I cut off piece of length 2 from the long blank (4).
After that we have blanks: 2 3 2 - two blanks coincide so one is eliminated: 2 3 they are different so we cut of blank with length 3. We have: 2 1. Cut again we have 1 1. Eliminate one blank and we get a result - 1.

You may try other ways to choose blank - but you will get same answer. So,
NO MATTER HOW TO CHOOSE 2 blanks.

Moral: What is not said does not matters ;)

By the way algorithm is very very easy... try to think

does anybody knows what is this fucking test 5?
if you know write the answer please:)

i've got tl on tets 7 cannot you say why?
i just calculate gcd of all numbers

Edited by author 26.04.2010 21:02

Edited by author 26.04.2010 21:03

#include <cstdlib>
#include <iostream>

//----------Funkcje----------------------------
long long nwd(long long a, long long b)
{
  if (b>=1) nwd(b, a%b);
  if (b=0) return a;
}


using namespace std;

//---------------Program------------------------------------
int main(int argc, char *argv[])
{
    int n;
    long long tab[1002]={0};
    //printf("Dla ilu liczb chcesz policzyc NWD?: \n");
    scanf("%d", &n);
    //printf("Podaj liczby: \n");
    for (int i=0;i<n;i++) scanf("%I64d", &tab[i]);
    long long wynik=tab[0];
    for (int i=0;i<n-1;i++) {
      wynik=nwd(tab[i],tab[i+1]);
      tab[i+1]=wynik;
    }

    //printf("NWD wynosi %I64d \n", wynik);
    printf("%I64d", wynik);

    system("PAUSE");
    return EXIT_SUCCESS;
}

#include <cstdlib>
#include <iostream>

//----------Funkcje----------------------------
long long nwd(long long a, long long b)
{
  if (b>=1) nwd(b, a%b);
  if (b=0) return a;
}


using namespace std;

//---------------Program------------------------------------
int main(int argc, char *argv[])
{
    int n;
    long long tab[1002]={0};
    //printf("Dla ilu liczb chcesz policzyc NWD?: \n");
    scanf("%d", &n);
    //printf("Podaj liczby: \n");
    for (int i=0;i<n;i++) scanf("%I64d", &tab[i]);
    long long wynik=tab[0];
    for (int i=0;i<n-1;i++) {
      wynik=nwd(tab[i],tab[i+1]);
      tab[i+1]=wynik;
    }

    //printf("NWD wynosi %I64d \n", wynik);
    printf("%I64d", wynik);

    system("PAUSE");
    return EXIT_SUCCESS;
}

Can anyone give me a test for which the answer is "impossible"?

#include

int gcd( int a, int b ){
  return b ? gcd( b, a%b ) : a;
}

int main () {
  int n, L, g=0;
  cin >> n;
  while( cin >> L ) g=gcd( L, g );
  cout << g << endl;
}

#include <stdio.h>
#include <stdlib.h>

int compare(const int *, const int *);
int gcd(int, int);

int main(void)
{
  int size = 0,counter = 0;
  scanf("%d",&size);
  int ls[size];
  int * pointer = ls;
  while(counter < size)
    {
      scanf("%d", pointer++);
      counter++;
    }
  qsort(ls, size, sizeof(int), compare);
  int x = ls[0];
  while(--counter > 0)
    {
      x = gcd(x, ls[counter]);
    }
  printf("%d", x);
  return 0;
}

int gcd(int small, int big)
{
  if (small == 0) return big;
  return gcd(big % small, small);
}

int compare(const int *p1, const int *p2)
{
  int x = *p1;
  int y = *p2;
  if(x < y)
    {
      return -1;
    }
  else
    {
      return 1;
    }
}

=====================

2904ea07-b44d-48c6-ba40-3b3f94228782
2904ea07-b44d-48c6-ba40-3b3f94228782(11): error: declaration may not appear after executable statement in block
    int ls[size];
    ^

2904ea07-b44d-48c6-ba40-3b3f94228782(11): error: expression must have a constant value
    int ls[size];
           ^

2904ea07-b44d-48c6-ba40-3b3f94228782(12): error: declaration may not appear after executable statement in block
    int * pointer = ls;
    ^

2904ea07-b44d-48c6-ba40-3b3f94228782(18): warning #167: argument of type "int (*)(const int *, const int *)" is incompatible with parameter of type "int (__cdecl *)(const void *, const void *)"
    qsort(ls, size, sizeof(int), compare);
                                 ^

2904ea07-b44d-48c6-ba40-3b3f94228782(19): error: declaration may not appear after executable statement in block
    int x = ls[0];
    ^

2904ea07-b44d-48c6-ba40-3b3f94228782(47): warning #1: last line of file ends without a newline
  }
   ^

compilation aborted for 2904ea07-b44d-48c6-ba40-3b3f94228782 (code 2)


=============

junmin@thinking ~/workspace/otros/ACM $ gcc --version
gcc (Gentoo 4.3.1-r1 p1.1) 4.3.1
Copyright (C) 2008 Free Software Foundation, Inc.
This is free software; see the source for copying conditions.  There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.

junmin@thinking ~/workspace/otros/ACM $ icc --version
icc (ICC) 10.1 20080801
Copyright (C) 1985-2008 Intel Corporation.  All rights reserved.


thanks!!!!

Edited by author 02.10.2008 23:11

Can anyone explain me the problem. Minimal
numbers????????????????? What on earth is that. I have gone
throuh he webboard but found nothing answering my question.
I just can't understand where the mistake is. In my
program? I don't know. But either the example or the
algorithm explained is absolutely wrong. Tell me please :)

If lengths of the blanks are different, then the piece of the most long blank is sawed off and its length must be equal to the length of the other blank. Then the algorithm is repeated from point 2.???

what does it mean

Edited by author 15.05.2005 19:59

Don't they mean planks ? :)

This is my programme,please tell me why I got WA.Thank you!

var n,m,m0,ys:longint;
procedure pd(a,b:longint);
  var t:longint;
  begin
  if a<b then begin
                  t:=b;
                  b:=a;
                  a:=t;
                end;
    while a mod b<>0 do
    begin
      a:=a mod b;
      t:=a;a:=b;b:=t;
    end;
    ys:=b;
  end;
procedure run;
  var i:integer;
  begin
    for i:=1 to n-1 do
    begin
      readln(m);
      pd(m0,m);
      m0:=ys;
    end;
    writeln(ys);
  end;
begin
  readln(n);
  read(m0);
  run;
end.

 who know please tell me ?
give me a sample programme is the best!

I'm quite sure that either the example or the problem statement is wrong....Is it possible to correct them?