//package timus;

import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.Reader;
import java.util.StringTokenizer;

public class p1423 {
    static char[] t;
    static char[] p;
    static int[] pi;
    static int N;

    public static void main(String[] args) throws FileNotFoundException {
        InputStream is = System.in;
        FastScanner sc = new FastScanner(new InputStreamReader(is));
        N = sc.nextInt();
        String tt = sc.next();
        tt += tt;
        t = tt.toCharArray();
        p = sc.next().toCharArray();
        pi = new int[N];
        System.out.println(KMPMatcher());
    }

    static int KMPMatcher() {
        computePrefixFunction();
        int equals = 0;
        for (int i = 0; i < N + N; i++) {
            while (equals > 0 && p[equals] != t[i])
                equals = pi[equals - 1];
            if (p[equals] == t[i])
                equals++;
            if (equals == N) {
                if (i == N - 1) {
                    return 0;
                } else {
                    return N + N - (i + 1);
                }
            }
        }
        return -1;
    }

    static void computePrefixFunction() {
        int k = 0;
        for (int q = 1; q < N; q++) {
            while (k > 0 && p[k] != p[q])
                k = pi[k - 1];
            if (p[k] == p[q])
                k++;
            pi[q] = k;
        }
    }

    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        FastScanner(Reader in) {
            br = new BufferedReader(in);
        }

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }
    }
}

"floor space one million square kilometers", so maximum number of slabs damaged by the fire is 1e12, but brute force solution (TL17) with ad-don "if(res>1e12)res=1e12" has WA14

#include <iostream>
#include <math.h>

using namespace std;

int main(){

    double x;
    int i;

    cin >> x;

    x = fmod (x, 7.0);

    printf("%.0f",x);

    system("pause");
    return 0;

}

Edited by author 18.08.2014 21:49

You don't need to calculate the topological order. If you first calculate the topological order, and then try to "verify" the study order, will complicate a simple problem.

Can anyone provide me 27th test case or tell me were am I wrong? I am using dynamic segment tree where in each node I store the count of doors in the current interval. Thanks in advance.

source code: https://ideone.com/PmG5Qb

Sorry for bad English

I had TLE on test 12 and realized that I used only simple recursion. Once I fixed it I got AC.

Edited by author 08.08.2015 14:46

Code:
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Globalization;

namespace _1601AntiCAPS
{
    class Program
    {
        static void Main(string[] args)
        {
            string input = Console.In.ReadToEnd();
            //string input;
            while (input != null)
            {
                Console.Write(input[0]);
                for (int i = 1; i < input.Length; i++)
                {

                    if (input[i] >= 'A' && input[i] <= 'Z')
                    {
                        int c = Convert.ToInt32(input[i]);
                        char character = (char)(c + 32);
                        string text = character.ToString();
                        Console.Write(text);
                    }
                    else
                        Console.Write(input[i]);
                }
            }
        }
    }
}

My program read string S from the input and create reverse string R from the source string. Then I search in cycle with prefix function maximum suffix which may be prefix on string S.substr(i, S.length()-i)+R   i - the number of the iteration.
Then I check if the getting string the palindrome, I remember then offset and length of the string. In the end print maximum substring. Got WA on 17 test.

Thanks on the future...:)

The test gives "wrong answer".
What is the test input?

A1A=10 1 10
For k=4,
10*(4^0)+1*(4^1)+10*(4^2)=174
174 is divisible by 3 so the answer for sample test case turns out to be 4.
What am I doing wrong? Thanks

What are answers for following comparisions?

'[]' like '[]'
'' like '[]'

As this problem's statement is really hard to interpret unambiguously, I've decided to make this attempt to explain my view on it, which seems to be the one that authors meant.
 Here’s the list of notes that make the problem more clear:

in template:
1. Brackets ‘[]’ have the highest priority. Find the leftmost ‘[‘, then find the first ‘]’ after it; you can assume that these two brackets strictly correspond to each other, i.e. can’t be considered to correspond to any other brackets. Then find the first opening bracket after current closing one and so on.
2. Such templates, as [] and [^] are incorrect and don’t happen in test cases.
3. Given an opening bracket [ and a corresponding closing bracket ], let’s consider the string S between them. As Vladimir Yakovlev already stated, special characters '[', '%' and '_' mean themselves when included in brackets. Now, let’s describe a boolean function F(c,s), that determines whether template s matches character c. "Del(s,k)" will denote a function, that returns the tail of s starting from position k+1 inclusively. We shall also need a boolean function G:

F:
If s[1]='^', Then F(c,s):=Not G(c,Del(s,1)), Else F(c,s):=G(c,s)

G:
If s[2]='–' and length(s)>=3 Then G(c,s):=G(c,Del(s,3)) Or (s[1]<=c<=s[3])
            Else G(c,s)=G(c,Del(s,1)) Or (s[1]=c)

To my mind, these rules do make the problem statement clear and have nothing to deal with posting Source code, so I hope moderators won’t delete this message. I also want to say thank you to the person posting under nick "I have answers to all your questions :)" for helping people on this problem. In fact, your posts helped me to understand it.

The problem unexpectedly dosn't contain very tricky tests.
Additionally to forum's examples I add the next:
1. Empty string corresponds with '%%%%%%%%' and so on and
with empty pattern also;
2. 'a'~'%a'
3. String may contain many "like" substrings . For defining
right position dublicable inner '-characters play key role;
4. Dp-method is veru useful, when we are going from ends of strings to their beginnings.

Edited by author 11.07.2007 17:24

'ab' like '[a]b'
'ab' like '%b'
'ab' like '%%b'
'ab' like '%%%b'
'ab' like '%%%b%'
'ab' like '%%%b%%'
'b' like '%%'

Edited by author 01.09.2010 00:37

What is the right answer for the given string?

i've got a wa on test 11st,could someone give me the test data?

Edited by author 26.02.2011 11:03

Edited by author 26.02.2011 11:03

It is the problem 1490 "Fire Circle" with the limitation of R equal to 10^9 instead of 10^5. If you have an efficient solution of problem 1490, try to solve this one.

Edited by author 02.08.2015 11:36

Can not figure out the right solution that passes the test