Try below test with empty line as a text:

--- start of input ---
world
#

--- end of input ---

Be careful, input in HEX can be presented as:
77 6F 72 6C 64 0D 0A
23 0D 0A
0D 0A

The answer is:
--- start of output ---

0
--- end of output ---

Be careful about an empty line before 0.

The following changes have been made to the problem:
- The checking program strictly checks that the initial formatting is preserved, for example it rejects outputs with additional empty lines before or after the text.
- Missing limitations added to the problem statements:
    - The words in the text to correct are no longer than 16 characters.
    - The lines in the text to correct are no longer than 80 characters.
    - The cumulative length of all lines doesn't exceed 10000 characters.
- New tests have been added.
- Time limit has been reduced to 0.5 sec.
- All solutions have been rejudged. 44% of authors have lost their accepted solutions.

#include <iostream>
#include <vector>
#include <string>
int corrections = 0;
bool words_match(const std::string& w1, const std::string& w2);
void process_word(const std::vector<std::string>& dict, const std::string& word);

int main(){
    std::vector<std::string> dictionary;
    std::string curr_dict_word;
    std::getline(std::cin, curr_dict_word);
    while(curr_dict_word != "#"){
        dictionary.push_back(curr_dict_word);
        std::getline(std::cin, curr_dict_word);
    }
    char c;
    std::string cword;
    while(std::cin.read(&c,1)){
        if('a'<=c && 'z'>=c){
            cword.insert(cword.end(), c);
            continue;
        }
        else if(!cword.empty()){
            process_word(dictionary, cword);
            cword = "";
        }
        std::cout << c;
    }
    if(!cword.empty()){
        process_word(dictionary, cword);
    }
    std::cout << corrections << std::endl;
    return 0;
}
bool words_match(const std::string& w1, const std::string& w2){
    if(w1.length() != w2.length()){
        return false;
    }
    bool mismatch = false;
    for(int i = 0; i < w1.length(); ++i){
        if(w1[i] != w2[i]){
            if(mismatch){
                return false;
            }
            mismatch = true;
        }
    }
    return mismatch;
}
void process_word(const std::vector<std::string>& dict, const std::string& word){
        for(int i = 0; i < dict.size(); ++i){
            if(words_match(dict[i], word)){
                std::cout << dict[i];
                ++corrections;
                return;
            }
        }
        std::cout << word;
}

It is said that text contains not more than 1000 words,
but limitations on length of words exist (in statement)
only for dictionary. It means that (technically)
Russian statement does not bound length of text
at all.

So, can you tweak problem statement a bit to make it more clear, please?

Edited by author 30.07.2014 20:05

#include <iostream>
#include <vector>
#include <string>
#include <cstring>
#include <string.h>
#include <fstream>

using namespace std;

struct our
{
    char a[10];
};

char temp[10];
our V[105];
int SizeV;

int main()
{
    freopen("output.txt","w",stdout);
    freopen("input.txt","r",stdin);
    bool doit = true;
    int SizeV = -1;
    while(doit)
        if(strlen(temp) == 1 && temp[0] == '#')
            doit = false;
        else
        {
            ++SizeV;
            for(int k = 0; k < strlen(temp); k++)
                V[SizeV].a[k] = temp[k];
            cin >> temp;
        }
    char a;
    char word[105];
    int fails = 0;
    string s;
    getline(cin,s);
    char r;
    while(getline(cin,s))
    {
        if(s.size() != 0)
        {
            int i = 0;
            char word[10];
            while(i < s.size())
            {
                int wordind = -1;
                while(s[i] >= 'a' && s[i] <= 'z')
                {
                    word[++wordind] = s[i];
                    i++;
                }
                bool done = false;
                int nuz = 0;
                for(int j = 0; j <= SizeV; j++)
                {
                    int error = 0;
                    if((wordind+1) == strlen(V[j].a))
                    {
                        for(int k = 0; k <= wordind; k++)
                            if(word[k] != V[j].a[k])
                            {
                                error++;
                                if(error > 1) break;
                            }
                        if(error == 1)
                        {
                            done = true;
                            nuz = j;
                            fails++;
                            break;
                        }
                    }
                }
                if(done)
                    for(int e = 0; e < strlen(V[nuz].a); e++)
                        cout << V[nuz].a[e];
                else
                    for(int e = 0; e <= wordind; e++)
                        cout << word[e];
                cout << s[i++];
            }
        }
        cout << endl;
    }
    cout << fails << endl;
    fclose(stdin);
    fclose(stdin);
    return 0;
}

Edited by author 09.07.2014 20:44

import java.io.*;

public class TaskTimus
{
    public static void main(String[] args) throws IOException
    {
        new TaskTimus().vvod();
    }

    String s = ""; String s1 = ""; String a[] = new String[120]; char c;
    int sch = -1, mistake = 0; String slovo = "";  int osh = 0; boolean bo = true;

    void vvod() throws IOException // INPUT THE DICTIONARY
    {
        BufferedReader r = new BufferedReader(new InputStreamReader(System.in));
        while (true)
        {
            slovo = r.readLine();
            if ("#".equals(slovo)) break;
            sch++;
            a[sch] = slovo;
        }
        prov();
    }

    void prov() throws IOException  //INPUT THE TEXT (FILE OR TIMUS CONSOLE)
    {

        boolean oj = System.getProperty("ONLINE_JUDGE") != null;
        Reader reader = oj ? new InputStreamReader(System.in) : new FileReader("C:\\test.txt");
        StreamTokenizer st = new StreamTokenizer(new BufferedReader(reader));
        st.eolIsSignificant(true);
        st.wordChars(32,47);
        st.wordChars(58,64);

        while (st.nextToken() != StreamTokenizer.TT_EOF)
        {
            if (st.ttype == StreamTokenizer.TT_WORD)
               s+=st.sval;

            if (st.ttype == StreamTokenizer.TT_EOL)
            {
                analysis();
                s = "";
            }
        }
        if (s!="") analysis();
        System.out.print(osh); // END OF INPUT, WRITE NUMBER OF MISTAKES
    }

    void analysis() throws IOException // CHECK
    {
        StringReader sr = new StringReader(s);
        for (int i = 0; i < s.length(); i++)
        {
                c = (char) sr.read();
                if (c <= 'z' && c >= 'a' || c <= 'Z' && c >= 'A') s1 += c;
                else mist();

        }
        if (s1!="") mist();
        System.out.println();
        sr.close();
    }

    void mist() throws IOException  // SEARCHING OF MISTAKE
    {
        for (int j = 0; j<=sch; j++)
        {
            if (a[j].length()==s1.length())
            {
                for (int k = 0; k<s1.length(); k++)
                {
                    if (a[j].charAt(k)!=s1.charAt(k)) mistake++;
                    if (mistake>1) { mistake = 0; break; }
                }
                if (mistake == 1)
                {
                   System.out.print(a[j]);
                   if (!(c <= 'z' && c >= 'a' || c <= 'Z' && c >= 'A')) System.out.print(c);
                   osh++; bo = false; mistake = 0; break;
                }
            }
        }
        if (bo) System.out.print(s1); if ( (!(c <= 'z' && c >= 'a' || c <= 'Z' && c >= 'A')) && bo) System.out.print(c);
        s1=""; bo = true; mistake = 0;
    }
}

Edited by author 08.07.2014 00:57

Please give me test. Thank you!

it seems like test#4 includes big latin letters, but it's impossible according problem statement "все слова записаны в нижнем регистре, слово может содержать только латинские буквы от 'a' до 'z' (то есть «one-seventh» — это два слова — «one» и «seventh»)."
admin, please check

It seems that in test case #7 there are more than 100 words in the vocabulary.  This contradicts with the problem description and should be fixed.

what is wrong?
var
    a:array[1..200] of string[8];
    b:string;
    ch:char;
    n,n1,n2,k,i,j,g,p,gg:integer;
begin
     n:=0; gg:=0;
     readln(b);
     p:=1;
     while b<>'#' do
     begin
          n:=n+1;
          a[n]:=b;
          readln(b);
     end;
     {$IFNDEF ONLINE_JUDGE}
     assign(input, 'input.txt');
     reset(input);
     assign(output, 'output.txt');
     rewrite(output);
     {$ENDIF}
     b:='';
     while not eof(input) do
     begin
          read(input,ch);
          if ch in ['a'..'z'] then b:=b+ch else
          begin
               k:=0;
               for i:=1 to n do
               begin
                    n1:=length(a[i]); n2:=length(b);
                    if n1=n2 then
                    begin
                         g:=0;
                         for j:=1 to n1 do if a[i][j]=b[j] then
                             begin
                                  g:=g+1;
                             end;
                         if g=n1-1 then
                                   begin
                                        b:=a[i];
                                        write(output,b);
                                        if p<>1 then write(output,ch);
                                        k:=1; gg:=gg+1;
                                   end;
                    end;
               end;
               if k=0 then write(output,b,ch);
               b:='';
               p:=2;
          end;
     end;
     writeln(output);
     write(output,gg);
     {$IFNDEF ONLINE_JUDGE}
     close(input);
     close(output);
     {$ENDIF}
end.

1. Do not check special characters such as '.' ','... only check characters 'a' - 'z' or 'A' - 'Z'
for example:
b
#
a. cd. .
output:
b. cd. .

2. Do not check equal strings
for example:
world
#
world world world

output:
world world world
0

3. Do not check strings having different length

Edited by author 12.07.2010 15:50

Edited by author 12.07.2010 15:51

A separator is a character which isn't a small latin letter.

I used two suffix array which contains sorted dictionary words

something need to take care:
1.if the readed word's lengh is within [1, 8]
2.if you found one in dictionary, make sure they have the same length and exact one letter different
3.The word readed in may contains only a single character

#include <stdio.h>
#include <string.h>

struct problem_1089
{
    char word[10];
    int len;
};

problem_1089 slovar[100] = {};
int n = 0;
bool flag = false;
char c;
char str[1001] = {};
int len = 0;
int count = 0;

void check()
{
    for (int i = 0; i < n; i++)
    {
        if (slovar[i].len == len)
        {
            int error = 0;
            for (int j = 0; j < len; j++)
            {
                if (str[j] != slovar[i].word[j]) error++;
                if (error > 1) break;
            }
            if (error == 1)
            {
                strcpy (str,slovar[i].word);
                count++;
                return;
            }
        }
    }
    return;
}

int main ()
{
    //freopen ("input.txt","r",stdin);
    //freopen ("output.txt","w",stdout);
    while (!flag)
    {
        gets(slovar[n++].word);
        slovar[n - 1].len = strlen (slovar[n - 1].word);
        if (strcmp(slovar[n - 1].word,"#") == 0)
        {
            flag = true;
            slovar[--n].word[0] = '\0';
            slovar[n].len = 0;
        }
    }
    bool perevod = false;
    while (scanf ("%c", &c) != EOF)
    {
        flag = false;
        if (c > 96 && c < 123)
            str[len++] = c;
        else
        {
            check();
            for (int i = 0; i < len; i++) printf ("%c", str[i]);
            if (c == '\n') perevod = true;
            else perevod = false;
            printf ("%c", c);
            len = 0;
        }

    }
    if (len > 0) perevod = false;
    check();
    for (int i = 0; i < len; i++) printf ("%c", str[i]);
    if (!perevod && !flag) printf ("\n%d", count);
    else printf ("%d", count);
    return 0;
}

if there is a empty line , don't stop your program.

1.There is one empty line in the end of text. You should not output
this line. But there are maybe some empty lines in the middle of
text, you should output these lines.

2.The lenght of words in the vocabulary may be larger than 8,the total
of words in the vocabulary may be exceed 100.

Читая различные обсуждения этой задачи, я нашел множество отклонений от условия... Так давайте верно переформулируем условие задачи!

On reading posts about the problem, i've found many mistakes in defination... Lets make the defination correct!

store all words in 2d array. the text in second 1d array.now read words one by one.compare each word with available match differing just by one letter.if words are mismatching by more than one letter or words belong to vocab then no need to change the text else change the text with word in vocab. best of luck!!

I solved this problem, but my program is wrong at 4 test.
Where is mistake?

var
l,k,j,m,n,mistake,o,sum,p,w:integer;  text1:string;
z,text:array[1..100]of string;
begin  o:=1;
repeat
inc(l);
readln(z[l]);
until z[l]='#';
dec(l);
repeat
inc(j);
readln(text[j]);
until text[j]=''; dec(j);
for k:=1 to l do
for m:=1 to j do
 begin
for n:=1 to length(text[m]) do
begin      text1:='';
while text[m][o] in ['a'..'z'] do
begin
text1:=text1+text[m][o]; inc(o);
end;
if length(text1)=length(z[k]) then
 begin
for w:=1 to length(z[k]) do
if text1[w]<>z[k][w] then inc(mistake);
if mistake=1 then
 begin
 inc(sum);
 for p:=1 to length(z[k]) do
  begin
  text[m][o-length(text1)+p-1]:=z[k][p];
  end;
 end;
mistake:=0;
end;  inc(o);
end;
o:=1;
(*----------------------*)
end;
for k:=1 to j do
writeln(text[k]);
writeln(sum);
end.

I've got WA#2 many times while i did not compare the length of tested word with the length of word in the vocabulary. So, i used strcmp() function and get AC in 0.001 sec with 124 KB of memory.