What is the test #21 ?

i used bresenham line drawing algorithm :)) together with gcd()

test:
 3
 A
 B
 ABP

 my ans: 3
         2 3 1

test
 5
 A
 B
 ABP
 AB
 AP

 my ans: 3
         2 4 3 5 1

test
 7
 A
 AP
 BP
 ABP
 P
 BP
 BP

 my ans: 3
         5 3 6 7 4 2 1


Edited by author 13.12.2015 17:36

I didn't find any case that my algorithm fails.

Please give me test cases for test #23, I see there are a lot of WA#23.

New anti-(N*N*M) test was added. TL was decreased to 1 second (this TL better corresponds to original 3 seconds in 2006). About 100 authors lost AC.

why do you always speak and write english?
aren't you Russian?
please, give the Russian version of this problem!!!!!

Can anyone give some test cases?

#include<bits/stdc++.h>
using namespace std;
int i, j, k, l, x, y, z, m, n, ans;
string first, second;
int dp[1010][1010][3][4], dir[1010][1010];

int make_way(int first_pile, int second_pile, char c, int state)
{

    //cout << first_pile << ' ' << second_pile << endl;
    if(first_pile == n && second_pile == n){
        if(state > 1) return 0;
        else return 1;
    }


    if(state > 1) return 0;

    int p = 0, q = 0, r = 0;
    char d;
    if(c == '0') d = '1';
    else d = '0';

    if(first_pile < n){
        if(first[first_pile] == c){
            p = make_way(first_pile + 1, second_pile, c, state + 1);
        }
        else{
            p = make_way(first_pile + 1, second_pile, d, 1);
        }
    }

    if(second_pile < n){
        if(second[second_pile] == c){
            q = make_way(first_pile , second_pile + 1, c, state + 1);
        }
        else{
            q = make_way(first_pile , second_pile + 1, d, 1);
        }
    }

    if(p == 1){
        dir[first_pile][second_pile] = 1;
    }
    else if(q == 1) dir[first_pile][second_pile] = 2;

    dp[first_pile][second_pile][c - '0'][state] = p | q;

    return dp[first_pile][second_pile][c - '0'][state];


}

int main()
{
    cin >> n >> first >> second;

    ans = make_way(0, 0, '0', 0);

    if(!ans){
        cout << "Impossible" ;
        return 0;
    }
    for(i = 0, j = 0; i < n || j < n; ){
        cout << dir[i][j];
        if(dir[i][j] == 1) i++;
        else j++;
    }

    return 0;
}

I guess input for test #10 is larger than 9.

According to my understanding of problem statement, output is always 1 or 2 or 3. The algorithm is pretty straightforward: if all the digits of input except first one, is 9 then print 3. If one among them is 8 and remaining is 9 then print 2, otherwise 1.

Any thoughts?

Hello,

I got WA#6 But for few days I'm unable to find a test that fails for my algorithm.
Does anyone has any hints? I would really appreciate any help.

Help me please!

i also got wa on 5 and then found out i considered room and floor both can be at most 100. but problem says floor is 100 and room can be at most 500.

My answers are fully identical to the given one. But I have WA#1.
Maybe, are there some format-specified features?

Try this:
1 1 -1

ans: 1 3

Why are not correct
12
4
6 3 1 1

1) (())(())(()) =>3
2) (())(()))))(() =>0
3) )))(()(()))((( =>2
4) )))))((()))))((((((( =>1
5) )()()()()()()()(())( =>9
6) ))(()( =>1

try
1 3
-80000 0 80000

I get WA 13 again and again and I can not find a mistake. My main logic looks like this:
for(j=0;j<n;j++)
{l=(x[j]-X)*(x[j]-X)+(y[j]-Y)*(y[j]-Y)+(z[j]-Z)*(z[j]-Z);
if (l-D<1e-10) D=l;}
printf("%0.6f\n",acos((2-D)/2.0)*r);

I believe it is correct..So can anybody help me with finding a bug, please? Does anyone have a same problem?

Please, somebody, give the 4th test