0100 the first '1' is in the position 2 or 3?

who has this in russian or
it will be better if somebody help me.
thanks

What you should consider is that, there will be some extra spaces at the end of ONE line. And then you'll get AC easily.

#include <fstream>
#include <stdio.h>
#include <iostream>
#include <string>
using namespace std;

int n;
string S;
int arr[1010] = {0};
char s[1010];

int main ()
{
    //freopen ("a.in", "r", stdin);
    //freopen ("a.out", "w", stdout);
    int i, j, k, t, sum;
    int tsum;
    scanf ("%d", &n);
    i = 0;
    S = "";
    char c;
    while (scanf ("%c", &c) == 1)
    {
        //cin>>S;
        //cout<<i<<" ";
        if (c == '\n' && S != "")
        {
            //S = "";
            /*for (j = 0; s[j] > 0; ++j)
            {
                if (s[j] == '0' || s[j] == '1')
                {
                    S += s[j];
                }
            }*/
            if (i > 0)
            {
                cout<<endl;
            }
            ++i;
            sum = 0;
            tsum = 0;
            if (S.length() == n)
            {
                for (j = 0; j < n; ++j)
                {
                    if (S[j] == '1')
                    {
                        sum += (j+1);
                    }
                }
                sum %= (n+1);
                if (sum == 0)
                {
                    cout<<S;
                    S = "";
                }
                else
                {
                    S[sum-1] = '0';
                    //S = "";
                    cout<<S;
                    S = "";
                }
            }
            else if (S.length() > n)
            {
                t = 0;
                for (j = 0; j < S.length(); ++j)
                {
                    if (S[j] == '1')
                    {
                        sum += (j+1);
                        ++t;
                    }
                    arr[j] = sum%(n+1);
                }
                sum %= (n+1);
                /*for (j = S.length()-1; j >= 0; --j)
                {
                    if (S[j] == '1')
                    {
                        tsum += (j+1);
                    }
                    arr[j][1] = tsum%(n+1);
                }*/
                if ((sum+n+1-t)%(n+1) == 0)
                {
                    S = S.substr (1, n);
                    cout<<S;
                    S = "";
                    continue;
                }
                t = 0;
                tsum = 0;
                for (j = S.length()-1; j > 0; --j)
                {
                    if ((tsum-t+arr[j-1])%(n+1) == 0)
                    {
                        for (k = 0; k < j; ++k)
                        {
                            cout<<S[k];
                        }
                        for (k = j+1; k < S.length(); ++k)
                        {
                            cout<<S[k];
                        }
                        //cout<<endl;
                        //continue;
                        S = "";
                        break;
                    }
                    if (S[j] == '1')
                    {
                        ++t;
                        tsum += (j+1);
                    }
                }
            }
            else
            {
                sum = 0, tsum = 0, t = 0;
                int add;
                for (j = 0; j < S.length(); ++j)
                {
                    if (S[j] == '1')
                    {
                        sum += (j+1);
                    }
                    arr[j] = sum%(n+1);
                }
                sum %= (n+1);
                //arr[n-1] = sum;
                if (sum == 0)
                {
                    cout<<S<<0;
                    S = "";
                    continue;
                }
                else if ((sum+n)%(n+1) == 0)
                {
                    cout<<S<<1;
                    S = "";
                    continue;
                }
                for (j = S.length()-1; j >= 0; --j)
                {
                    if ((arr[j]+tsum+t)%(n+1) == 0)
                    {
                        for (k = 0; k <= j; ++k)
                        {
                            cout<<S[k];
                        }
                        cout<<0;
                        for (k = j+1; k < S.length(); ++k)
                        {
                            cout<<S[k];
                        }
                        //cout<<endl;
                        S = "";
                        break;
                    }
                    if ((arr[j]+tsum+t+j+1)%(n+1) == 0)
                    {
                        for (k = 0; k <= j; ++k)
                        {
                            cout<<S[k];
                        }
                        cout<<1;
                        for (k = j+1; k < S.length(); ++k)
                        {
                            cout<<S[k];
                        }
                        S = "";
                        //cout<<endl;
                        break;
                    }
                    if (S[j] == '1')
                    {
                        ++t;
                        tsum += (j+1);
                    }
                }
            }
            S = "";
        }
        else
        {
            if (c == '0' || c == '1')
            {
                S += c;
            }
        }
    }
    return 0;
}

For example program must work some:
4 <INPUT
0000 <INPUT
0000 >OUTPUT
011  <INPUT
0110 >OUTPUT
1011 <INPUT
1001 >OUTPUT
11011<INPUT
1111 >OUTPUT

using EOF allways in INPUT

Happy...

Can you give me some tests?

Edited by author 16.03.2007 06:48

My WA Reason was the extra line break at the end of output! Be Careful!!!

My solution take WA in Test # 1
If someone can help me please, write is what my mistake.
code:

[code deleted]

Edited by moderator 13.02.2007 20:49

Giv my please Test #1?
Can I not it is correct data read.

HELP PLEASE??? ANYBODY???

For example:
4
1010

In statement was written:
"3. A symbol (0 or 1) is inserted at any position."

How many symbols can be inserted at any position?

 There is nothing else in the file, except maybe for some extra spaces or line breaks.
What das it mean?
Maybe?
#input 1
4
000


000
--------
OR
#input 2
6
000 00
111111

000111

Who solve this problem please tell how read input.
I use
i=scanf("%s",str1);
    if(i==EOF){
        size=1;
        return ;
    }
//while size >1
But always get WA1.
Sorry for my English.

I know that it is stupid to read all strings and then process them, instead of reading one - processing it, reading next, processing it and so on.
But I coded as wrote before(read all and then solve)

I created int r[1024][1024], because in the task there is written that "There are no more than 1000 words in the file", and size of each is no more than 1001(initially is 1000, but with one inserted symbol...). I got Crash(access violation), I guessed that maybe I addressed inaccessible memory.
I updated my code and replaced it by this:

char r[1024][1024];

    string s;
    cin>>n;

    m = 0;
    cin>>s;
    while (!(cin.eof())){
        if (m > 1000)
            continue;
        if (s.size()>n+1)
            return;
        if (s.size()<n-1)
            return;
        strcpy(r[m++],s.c_str());
        len[m-1] = strlen(r[m-1]);
        cin>>s;
    }

and got TL test 3. So it was, I think, because of
        if (m > 1000)
            continue;

then I made int r[3000][1024] and got AC ))
why? maybe there is more than 1000 words in the file?

How do I number the positions?

[code deleted]

Could anyone help me?

Edited by moderator 08.07.2006 22:33

  I have WA for test#1 but i verified the problem with all possible words of lengths between 2 and 20 and it returned correct answers.
  Also I have taken into account the possibility of entering words like:
 0000
 0001

 100


 101

or
00 0 0
10  0

but it still says WA.

email : sraduvictor@yahoo.com

I tried to leave main cycle when the input is null-string, but I got "Output Limit Exceeded". More over, the problem says that there may be some line breaks.
So what the point?

I output the answer for a consecutive binary word right after the input...

What is "Output Limit" ?