Precalculate pythagorean tripples

The statement is poor. It says M*N, which is usually considered as M in x-axis and N in y-axis.
But actually, for M=2,N=3 they mean
xxx
xxx
(That can be infered from "The next M lines contain a matrix with nonnegative integer elements")
And we can't understand that from Sample because it's a square!

***  So,if you always get WA2,try to swap m,n.  ***
I changed scanf("%d %d",&m,&n) to scanf("%d %d",&n,&m) and got AC.
Sorry for my poor English.

Edited by author 15.08.2009 11:30

Edited by author 15.08.2009 11:31

Edited by author 08.06.2009 19:23

I've seen lots of guys crashed on test 2. Could anyone provide a hint for this test?

1. Even number of changes does not change the color. For eg : B -> W -> B
2. Odd number of changes equals to 1, B->W and vice versa

Edited by author 01.10.2010 12:24

9 9
WWWWWWWWW
WWWWWWWWW
WWWWWWWWW
WWWWWWWWW
WWWWWWWWW
WWWWWWWWW
WWWWWWWWW
WWWWWWWWW
WWWWWWWWW
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0

correct answer
WBWWBWWBW
BWWWBWWWB
WWWWBWWWW
WWWWBWWWW
BBBBBBBBB
WWWWBWWWW
WWWWBWWWW
BWWWBWWWB
WBWWBWWBW

Here is my source :

#include <stdio.h>

int n,m;
long y[60][60];
char in[60][60],x[60][60],sel[3000];

void read_data()
 {
  int i,j;
  FILE *f=stdin;
  fscanf(f,"%d%d",&m,&n);
  fscanf(f,"%c",&in[0][0]);
  for(i=0;i<m;i++)
   fgets(in[i],100,f);
  for(i=0;i<m;i++)
   for(j=0;j<n;j++)
    fscanf(f,"%ld",&y[i][j]);
 }

void init()
 {
  int i;
  for(i=0;i<=50;i++)
   sel[i*i]=1; //sel[i]== 1 if i==(X*X)
               //         0 else
 }

void expand(int a,int b,int nr)
 {
  int i,j;
  for(i=0;i<m;i++)
   for(j=0;j<n;j++)
    if(j==b || i==a || sel[((i-a)*(i-a)+(j-b)*(j-b))])
     x[i][j]=x[i][j] ^ nr;
 }

void solve()
 {
  int i,j;
  init();
  for(i=0;i<m;i++)
   for(j=0;j<n;j++)
    if(y[i][j])expand(i,j,y[i][j]%2);
 }

void print()
 {
  int i,j;
  FILE *f=stdout;
  for(i=0;i<m;i++)
   {
    for(j=0;j<n;j++)
     if(x[i][j])
      if(in[i][j]=='B')fprintf(f,"W");
      else fprintf(f,"B");
     else fprintf(f,"%c",in[i][j]);
    fprintf(f,"\n");
   }
 }

void main()
 {
  read_data();
  solve();
  print();
 }

Why the cell [6][4] is 'W' in the output?
We changed this cell 138 times.
I think it must be 'B' as in input.
Can U explain me?

There are N rows And M columns.

I have a WA on the test 2. Who nows it, please, help me.

My program:

var  i,j,n,m,x:longint;
     s:array[1..50,1..50]of string[1];
     a:array[1..50,1..50]of integer;

 procedure redata;
  begin
    fillchar(a,sizeof(a),0);
   readln(n,m);
   for i:=1 to n do begin
   for j:=1 to m do
    read(s[i,j]);
    readln;   end;
   for i:=1 to n do begin
   for j:=1 to m do begin
    read(x);
    if x mod 2=1 then a[i,j]:=1; end;
    readln;   end;


  end;

  procedure zamena;
   var k,x,y:integer;
   begin
    for k:=1 to m do
     if s[i,k]='W' then s[i,k]:='B' else s[i,k]:='W';
    for k:=1 to n do
     if s[k,j]='W' then s[k,j]:='B' else s[k,j]:='W';

    if s[i,j]='W' then s[i,j]:='B' else s[i,j]:='W';

    for X:=1 to 50 do
     for Y:=X to 50 do

     if frac(sqrt((X*X)+(Y*Y)))=0 then begin

     if (i-x>0)and(j-y>0) then BEGIN
     if s[i-x,j-y]='W' then s[i-x,j-y]:='B' else s[i-x,j-y]:='W';END;

     if (i-y>0)and(j-x>0) then BEGIN
     if s[i-y,j-x]='W' then s[i-y,j-x]:='B' else s[i-y,j-x]:='W';END;

     if (i-y>0)and(j+x<=N) then BEGIN
     if s[i-y,j+x]='W' then s[i-y,j+x]:='B' else s[i-y,j+x]:='W';END;

     if (i+y<=M)and(j-x>0) then BEGIN
     if s[i+y,j-x]='W' then s[i+y,j-x]:='B' else s[i+y,j-x]:='W';END;

     if (i-x>0)and(j+y<=N) then BEGIN
     if s[i-x,j+y]='W' then s[i-x,j+y]:='B' else s[i-x,j+y]:='W';END;

     if (i+x<M)and(j-y>0) then  BEGIN
     if s[i+x,j-y]='W' then s[i+x,j-y]:='B' else s[i+x,j-y]:='W';END;

     if (i+x<=M)and(j+y<=N) then BEGIN
     if s[i+x,j+y]='W' then s[i+x,j+y]:='B' else s[i+x,j+y]:='W';END;

     if (i+y<=M)and(j+x<=M) then BEGIN
     if s[i+y,j+x]='W' then s[i+y,j+x]:='B' else s[i+y,j+x]:='W';END;

                           end;
   end;

 procedure main;
  begin
   for i:=1 to n do
   for j:=1 to m do
    if a[i,j]=1 then zamena;
  end;

 procedure outdata;
  begin
   for i:=1 to n do begin
   for j:=1 to m do
   write(s[i,j]);
   writeln;  end;

  end;

 begin
  redata;
  main;
  outdata;
 end.

First i Think that it means horisontal and vertical num? but
when i have written my program it did not pass that test!
What mean integer num of meter?

const maxn=50;
type hl=array[1..maxn] of 0..1;
var data:array[1..maxn,1..maxn] of longint;
    f:array[1..maxn,1..maxn] of 0..1;
    h,l:hl;
    i,j,k,n,m:longint;
    color:array[1..maxn,1..maxn] of char;

function min(r,t:longint):longint;
  begin
    if r<t then min:=r else min:=t
  end;

procedure make(x,y:longint);
  begin
    for k:=min((m-i) div x,(n-j) div y) downto 1 do

      f[i+k*x,j+k*y]:=1-f[i+k*x,j+k*y];

    for k:=min((m-i) div x,(j-1) div y) downto 1 do

      f[i+k*x,j-k*y]:=1-f[i+k*x,j-k*y];

    for k:=min((i-1) div x,(j-1) div y) downto 1 do

      f[i-k*x,j-k*y]:=1-f[i-k*x,j-k*y];

    for k:=min((i-1) div x,(n-j) div y) downto 1 do

      f[i-k*x,j+k*y]:=1-f[i-k*x,j+k*y]

  end;

begin
  assign(input,'input.in'); reset(input);
  readln(m,n);
  for i:=1 to m do
    begin
      for j:=1 to n do read(color[i,j]);
      readln
    end;
  for i:=1 to m do for j:=1 to n do read(data[i,j]);
  close(input);

  fillchar(h,sizeof(h),0); l:=h;
  fillchar(f,sizeof(f),0);

  for i:=1 to m do
    for j:=1 to n do
        if odd(data[i,j]) then begin
          h[i]:=1-h[i];
          l[j]:=1-l[j];
          make(3,4); make(4,3); make(5,12); make(12,5)
        end;
  for i:=1 to m do
    begin
      for j:=1 to n do
        if color[i,j]='W' then if odd(h[i]+l[j]+f[i,j]+data[i,j])
then write('B') else write('W')
          else if odd(h[i]+l[j]+f[i,j]+data[i,j]) then write('W')
else write('B');
      writeln
    end
end.

My pro
  #include<stdio.h>
#include<math.h>
main()
{
  int m,n,i,j,b[50][50],data,k,p,q,r,h,t[50][50];
  char a[50][50],c;
  scanf("%d%d",&m,&n);
  c=getchar();
  for(i=0;i<50;i++)
    for(j=0;j<50;j++)
      t[i][j]=0;
  for(i=0;i<m;i++)
   {
     for(j=0;j<n;j++)
     {
    c=getchar();
    if(c=='W')
      a[i][j]=0;
    else a[i][j]=1;
     }
    c=getchar();
   }
  for(i=0;i<m;i++)
    for(j=0;j<n;j++)
      {
    scanf("%d",&data);
    b[i][j]=data%2;
       }
 for(i=0;i<m;i++)
  for(j=0;j<n;j++)
    {
      if(b[i][j]==1)
    {
    /*  for(k=0;k<m;k++)
        {
         if(a[k][j]==1)
           a[k][j]=0;
         else a[k][j]=1;
         }
       for(k=0;k<n;k++)
        {
          if(a[i][k]==1)
        a[i][k]=0;
           else a[i][k]=1;
         }  */
      /* if(a[i][j]==1)
         a[i][j]=0;
       else a[i][j]=1;*/
    for(p=0;p<m;p++)
      for(q=0;q<n;q++)
        {
          if(p==i)
        t[p][q]++;
           else if(q==j)
           t[p][q]++;
          else{
          r=(p-i)*(p-i)+(q-j)*(q-j);
          h=sqrt(r);
          if(r==h*h)
           {
        t[p][q]++;

        }
        }     }
       }
     }
for(i=0;i<m;i++)
  for(j=0;j<n;j++)
    {
      t[i][j]=t[i][j]%2;
      if(t[i][j]==1)
     {
       if(a[i][j]==0)
         a[i][j]=1;
        else a[i][j]=0;
       }
      }

for(i=0;i<m-1;i++)
  {
   for(j=0;j<n;j++)
    {
     if(a[i][j]==0)
       printf("W");
      else printf("B");
     }
   printf("\n");
  }
for(j=0;j<n;j++)
  {
    if(a[i][j]==0)
      printf("W");
      else printf("B");
  }
}