If you have CRASH, may be, will help for you to make columns of a global variable and to clean from definitions of procedures. I got AC...

If you get WA2 try this one:
Input:
3
3 2
2 1
1 -1
3
1 2
3 2
1 3
Output:
1
2
1

Also if you get WA12 pay attention for: "All identifiers lie between 0 and 40000."

#include <iostream>
#include <iomanip>
#include<math.h>
#include<vector>
using namespace std;
int  main()
{
    cout<<setiosflags(ios::fixed)<<setprecision(4);
    vector<double>v;
    double a;
    for(int i=0;!cin.eof();i++ )
    {
        cin>>a;
        v.push_back(sqrt(a));
    }
    for(int j=v.size()-2;j>=0;j--)
    {cout<<v[j]<<"\n";}
    return 0;
}

How to control the output precision?What does "to the acuracy of 10-6?

 var
  i,j,k,n,m,tot:longint;
  ans,x1,y1,x2,y2,d1,d2,d3,x,y,angle1,angle2,t,p1,p2:double;

 function min(xx,yy:double):double;
  begin
   if xx>yy then exit(yy);
   exit(xx);
  end;

 begin
  {assign(input,'1215.in');reset(input);}
  read(x,y,n);
  read(x1,y1);
  p1:=x1;
  p2:=y1;
  d1:=sqrt(sqr(x-x1)+sqr(y-y1));
  ans:=999999999;
  tot:=0;
  for i:=2 to n do
   begin
    read(x2,y2);
    if (x1<x)and(x2>=x)and((y1<y)or(y2<y))
     then
      inc(tot);
    d2:=sqrt(sqr(x-x2)+sqr(y-y2));
    d3:=sqrt(sqr(x1-x2)+sqr(y1-y2));
    angle1:=(d2*d2+d3*d3-d1*d1)/(2*d2*d3);
    angle2:=(d1*d1+d3*d3-d2*d2)/(2*d1*d3);
    if (angle1>=0)and(angle2>=0)
     then
      begin
       if x1=x2
        then
         t:=abs(x1-x)
        else
         t:=((y1-y2)/(x1-x2)*x-y+y1+(y2-y1)/(x1-x2)*x1)/(sqrt(sqr((y1-y2)/(x1-x2))+1));
       t:=abs(t);
       ans:=min(ans,t);
      end
     else
      begin
       t:=min(d1,d2);
       ans:=min(ans,t);
      end;
    d1:=d2;
    x1:=x2;
    y1:=y2;
   end;
  x2:=p1;
  y2:=p2;
  if (x1<x)and(x2>=x)and((y1<y)or(y2<y))
     then
      inc(tot);
    d2:=sqrt(sqr(x-x2)+sqr(y-y2));
    d3:=sqrt(sqr(x1-x2)+sqr(y1-y2));
    angle1:=(d2*d2+d3*d3-d1*d1)/(2*d2*d3);
    angle2:=(d1*d1+d3*d3-d2*d2)/(2*d1*d3);
    if (angle1>=0)and(angle2>=0)
     then
      begin
       if x1=x2
        then
         t:=abs(x1-x)
        else
         t:=((y1-y2)/(x1-x2)*x-y+y1+(y2-y1)/(x1-x2)*x1)/(sqrt(sqr((y1-y2)/(x1-x2))+1));
       t:=abs(t);
       ans:=min(ans,t);
      end
     else
      begin
       t:=min(d1,d2);
       ans:=min(ans,t);
      end;
  if odd(tot)
   then
    ans:=0;
  writeln(ans*2:0:3);
 end.

just check your program on these tests
----------
4 3
1 5 8 10
answer:
1 8
5 10
wrong answer:
1 5
8
10
----------
4 2
3 5 7 10
answer:
5 10
3 7
wrong answer:
7 10
5
3

I know AC solution that answers 5 on this test instead of 3:
/////////////start of test
6






////////////end of test

6
[1 space]
[1 space]
[2 spaces]
[2 spaces]
[3 spaces]
[3 spaces]
[EOF]

Also this test should be answered 0 instead of 1.
2
 MEXX
MEXX

I gues this tests suit the condition "The brands are the strings of Latin letters and blanks." and may be added.

Thank you.

It works like this:
f1=1*f1+0*f2;
f2=0*f1+1*f2;
f3=1*f1+1*f2;
f4=1*f1+2*f2;
f5=2*f1+3*f2;
f6=3*f1+5*f2;
...^....^
...|....These coefficients are in the array c2
...These coefficients are in the array c1

----PROG BELOW----
program ural1133;
const
  maxn=2001;
var
  c1,c2:array[1..maxn]of extended;
  x,y,n,i:integer;
  fx,fy,f1,f2:longint;
function min(a,b:integer):integer;
  begin
    if a<b then min:=a else min:=b;
  end;
function max(a,b:integer):integer;
  begin
    if a>b then max:=a else max:=b;
  end;
begin
  readln(x,fx,y,fy,n);
  i:=min(min(x,y),n);
  if i<=0 then begin
    i:=1-i;
    inc(x,i);inc(y,i);inc(n,i);
  end;

  c1[1]:=1;c2[1]:=0;
  c1[2]:=0;c2[2]:=1;
  for i:=3 to max(max(x,y),n) do begin
    c1[i]:=c1[i-2]+c1[i-1];
    c2[i]:=c2[i-2]+c2[i-1];
  end;

  f1:=round((fx*c2[y]-fy*c2[x])/(c1[x]*c2[y]-c1[y]*c2[x]));
  f2:=round((fx*c1[y]-fy*c1[x])/(c2[x]*c1[y]-c2[y]*c1[x]));
  writeln(round(c1[n]*f1+c2[n]*f2));
end.

Test
1 -2000000000 2 2000000000 3

is correct?

Answer: 0

My first code:

#include <stdio.h>
int n;
long long f[501][501], g[501][501], res;
void xxx(void)
{
     for (int i=0; i<=n; i++) f[0][i] = 1;
     for (int i=1; i<35 && i<=n; i++) {
         for (int j=0; j<=n; j++)
             for (int k=0; k<=n; k++) {
                 g[j][k] = f[j][k]; f[j][k] = 0;
             }
         for (int j=1; j<=n; j++) {
             for (int k=1; k<=j; k++) f[j][k] = f[j][k - 1] + g[j - k][k - 1];
             for (int k=j+1; k<=n; k++) f[j][k] = f[j][j];
         }
         res += f[n][n];
     }
}
int main(void)
{
    scanf("%d", &n);
    xxx();
    printf("%lld\n", --res);
    return 0;
}
WA for #6.

My second code:

#include <stdio.h>
int n;
__int64 f[501][501], g[501][501], res;
void xxx(void)
{
     for (int i=0; i<=n; i++) f[0][i] = 1;
     for (int i=1; i<35 && i<=n; i++) {
         for (int j=0; j<=n; j++)
             for (int k=0; k<=n; k++) {
                 g[j][k] = f[j][k]; f[j][k] = 0;
             }
         for (int j=1; j<=n; j++) {
             for (int k=1; k<=j; k++) f[j][k] = f[j][k - 1] + g[j - k][k - 1];
             for (int k=j+1; k<=n; k++) f[j][k] = f[j][j];
         }
         res += f[n][n];
     }
}
int main(void)
{
    scanf("%d", &n);
    xxx();
    printf("%I64d\n", --res);
    return 0;
}
AC.

20 7
D 9
D 7
D 6
D 1
D 2
D 3
L 2
The answer should be 4.

During 3 days I couldn’t find appropriate way
to work with big binomials having lost of order
or overflow. Simple and clever routine was found and
gave satisfaction.

You can see two program with one difference - type. What wrong with first program?

define _CRT_SECURE_NO_DEPRECATE
#define _USE_MATH_DEFINES

#include <utility>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstdlib>
#include <string>
#include <cstring>
#include <queue>

using namespace std;

long long n, m, y, x;
bool yes = 0;
int main()
{
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    scanf("%lld%lld%lld",&n,&m,&y);
    for (int i = 0; i < m; ++i)
    {
        x = i;
        if (n == 0) x = 1;
        for (int j = 2; j <= n; ++j)
            x = (x*i)%m;
        if (x%m == y)
        {
            yes = 1;
            printf("%lld", i);
            printf(" ");
        }
    }
    if (!yes) printf("-1");
    return 0;
}

#define _CRT_SECURE_NO_DEPRECATE
#define _USE_MATH_DEFINES

#include <utility>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstdlib>
#include <string>
#include <cstring>
#include <queue>

using namespace std;

int n, m, y, x;
bool yes = 0;
int main()
{
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    scanf("%d%d%d",&n,&m,&y);
    for (int i = 0; i < m; ++i)
    {
        x = i;
        if (n == 0) x = 1;
        for (int j = 2; j <= n; ++j)
            x = (x*i)%m;
        if (x%m == y)
        {
            yes = 1;
            printf("%d", i);
            printf(" ");
        }
    }
    if (!yes) printf("-1");
    return 0;
}

program p1423;
var
  p:array[0..250000]of longint;
  a,b:array[0..250000]of char;
  i,j,k,n,m:longint;
begin
  readln(n);
  for i:=1 to n do
    read(b[i]);
  readln;
  for i:=1 to n do
    read(a[i]);
  readln;
  p[1]:=0;
  j:=0;
  for i:=2 to n do
    begin
      while(j>0)and(b[j+1]<>b[i])do
        j:=p[j];
      if b[j+1]=b[i]
        then j:=j+1;
      p[i]:=j;
    end;
  j:=0;
  for i:=1 to n do
    begin
      while(j>0)and(b[j+1]<>a[i])do
        j:=p[j];
      if b[j+1]=a[i]
        then j:=j+1;
    end;
  if j=n
    then begin
      writeln(0);
      halt;
    end;
  if j<>0
    then writeln(n-j)
    else writeln(-1);
end.

Thanks for helping!
Sorry for my English...

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