My program gives  right answers for all your tests, but I got WA7. Please help.

Thank you!

Can you give me test 3

Edited by author 22.02.2007 17:21

When I submit the program with the comparison function shown below, I get WA 27. But if I use a similar comparison function as bruce described in the other topic, i get AC. What's weird is that both result in the same answer (I assert-ed this), but maybe different permutations. So unless my evaluation function (also shown below) is wrong, the judge does not accept all valid permutations.

ll calc() {
    ll ca=0,cb=0,cc=0;
    REP(i,n) { ca+=x[i].a; cb=max(ca,cb)+x[i].b; cc=max(cb,cc)+x[i].c; }
    return cc;
}



typedef struct cmp2 {
    bool operator() (const X &x,const X &y) {
        return min(y.a+y.b,x.b+x.c)>min(x.a+x.b,y.b+y.c);
    }
} cmp2;

Edited by author 19.02.2007 05:37

I carefully wrote my solution to not put any extraneous spaces at the end of the second line and it came back WA #1 every time during the contest. If I now put a space at the end of the line, it comes back as WA #12.

Please give me some more test cases!

<code deleted>

Edited by author 20.02.2007 19:37

Please, give me same test. I hava WA on test 7.

why example 1:
1 1 1
uuuur

I think it should be 0!!

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.

What does it mean?
And why the sample result #1 is 1?

#include <iostream.h>
#include <stdio.h>
#include <string.h>
int p[4]={-1,-1,-1,-1};
struct student
{
    char name [256];
};
int getfac(char* fn);
int main ()
{
    int n,i,j,id;
    struct student stud[4][1000];
    char* fc[4]={"Slytherin","Hufflepuff","Gryffindor","Ravenclaw"};
    char tmp1[256];
    char tmp2[256];
    scanf("%i", &n);
    for(i=0;i<n;i++)
    {
        //scanf("%s\n%s\n",tmp1,tmp2);
        //printf("%s\t-\t%s",tmp1,tmp2);
        cin.getline(tmp1,256);

        //cout<<tmp1<<endl;
        cin.getline(tmp2,256);
        //cout<<tmp2<<endl;
        id=getfac(tmp2);
        strcpy(stud[id][p[id]].name,tmp1);


    }
    for(i=0;i<4;i++)
    {
        if(p[i]>-1)
            cout<<fc[i]<<":\n";
        for(j=0;j<p[i]+1;j++)
        {
            cout<<stud[i][j].name<<"\n";
        }
        cout<<"\n";
    }
    /*char x[256];
    cin.getline (x,256);
    cout<<x<<endl;*/
    return 0;
}

int getfac(char* fn)
{
    if(!strcmp(fn, "Slytherin"))
    {
        p[0]++;
        return 0;
    }
        if(!strcmp(fn, "Hufflepuff"))
    {
        p[1]++;
        return 1;
    }
        if(!strcmp(fn, "Gryffindor"))
    {
        p[2]++;
        return 2;
    }
            if(!strcmp(fn, "Ravenclaw"))
    {
        p[3]++;
        return 3;
    }
    return -1;
}

If A = "010" and B = "010", then A + B = "100" ?

There are two algorithms of determining a bridge in the graph: simple and advanced. Simple algorithm O(n^2) works more than 1s if written in Java, but less than 1s if written in C++.

I suggest to lower time limit to 0.5s, after that only advanced solutions O(n), (n=number of islands) will be accepted.

I got Output Limit on Test#3
This Is My CODE:

#include<iostream>
using namespace std;


char stack[50000][14];
int pos=0;
bool ii=true;

void push(char a[4],char b[9])
{
    int i;
    int l=strlen(a);
    for(i=0;i<l;i++)
        stack[pos][i]=a[i];
    stack[pos][i]='#';
    i++;
    int ll=l+strlen(b)+1;
    for(;i<ll;i++)
        stack[pos][i]=b[i-l-1];
    pos++;
}
void pop(char a[4])
{


    int i;
    int j;
    for(i=pos-1;i>=0;i--)
    {
        ii=true;
        j=0;
        if(stack[i][0]!='~')
        {
        while(stack[i][j]!='#')
        {
            if(a[j]!=stack[i][j])
            {
                ii=false;
                break;
            }
            j++;
        }
        j++;


        if(ii)
        {
            int sl=strlen(stack[i]);
            for(;j<sl;j++)
            {
                cout<<stack[i][j];
            }
            cout<<endl;




            stack[i][0]='~';
        return;
        }
        }


    }

}

int main()
{
    char a[4],b[9];
    char tmp[4];
    char cc;
    int n;
    cin>>n;
    for(;n>=1;n--)
    {
    tmp[0]='~';tmp[1]='~';tmp[2]='~';tmp[3]='~';
    cin>>tmp;
    if(tmp[1]=='U')
    {
        cin>>a>>b;
        push(a,b);
    }
    else
    {
        cin>>a;
        pop(a);
    }
    }




    return 0;
}

Edited by author 21.02.2007 00:33

Who can explain me how to solve this problem?

I didn't understand the question!!! exactly what must our output 123<-456<-7890 <<<<<<<<<<<<<<<<to order <<<<<<<<<<<<<<<<<computers and ...

task 1014
input output
1     1
0     10
2     2
10    25

test 8:
?     ?

??????????????????

Edited by author 20.02.2007 12:41