time 0.078
memory 3 350 kb.

I use triplex tree.
It is very interesting problem.

#include <stdio.h>
#include <iostream>
using namespace std;
int n, m, segRez[3004];
int segm1[3004];
char segm2[6005];
int main()
{
    cin >> n >> m;
    for(int i = 0; i < n; ++i){
        gets_s(segm2);
        for(int j = 0; j < m; ++j){
            segRez[j] = segm1[j] + int(segm2[j<<1] - 48);
            segm1[j] = int(segm2[j<<1] - 48);
            if(j - 1 >= 0 && ((segRez[j - 1]|segRez[j])) == 3){
                cout << "No";
                return 0;
            }
        }
    }
    cout << "Yes";
}

Here is my java code :


import java.util.Scanner;
public class Example{
    public static void main(String []args){
        Scanner in = new Scanner(System.in);
        int number1,number2,number3;
        int [] array;
        int position=0;
        boolean check=false;
        int sth;
        number1=in.nextInt();
        number2=in.nextInt();
        number3=in.nextInt();
        array = new int[number2];
        for(int i=0;i<=number2-1;i++){
            sth=(int)Math.pow(i,number1);
            if((sth-number3)%number2==0){
                array[position]=i;
                position++;
                check=true;
            }else if(i==number2-1 && check==false){
                System.out.println(-1);
                System.exit(0);
            }

        }
        for(int p=0;p<position-1;p++){
            for(int j=p+1;j<position;j++){
                if(array[p]>array[j]){
                    int t=array[p];
                    array[p]=array[j];
                    array[j]=t;
                }
            }
        }
        for(int i=0;i<position;i++){
            System.out.print(array[i] + " ");
        }
        System.out.println();
    }
}

I get an error on test#3 .
And on my C++ solution i get an error on test#6.



#include<iostream>
#include<cmath>
using namespace std;
int main(void){
unsigned short int N,M,Y;
cin>>N>>M>>Y;
unsigned short int X[M];
unsigned short int position=0;
bool some_check=false;
int sth;
for(int i=0;i<=M-1;i++){
    sth=pow(i,N);
    if((sth-Y)%M==0){
        X[position]=i;
        position++;
        some_check=true;
    }
    else if(i==M-1 && some_check==false){
        cout<<-1<<endl;
        return 0;
    }
}
for(unsigned short int i=0;i<position-1;i++){
    for(unsigned short int j=i+1;j<position;j++){
        if(X[i]>X[j]){
            swap(X[i],X[j]);
        }
    }
}
for(unsigned short int i=0;i<position;i++){
    cout<<X[i]<<" ";
}
cout<<endl;
return 0;
}


Please,any ideas ???

int main()
{
    int n;
    string s1, s2;
    cin >> n >> s1 >> s2;
    s1 += s1;
    int pos = (int) s1.find(s2);
    switch ( pos )
    {
    case -1:
        cout << -1;
        break;
    case 0:
        cout << 0;
        break;
    default:
        cout << n-pos;
    }
    return 0;
}

¿Any idea of why a code can pass all the tests except the test 6 ? I prove my code with this imput, and it works fine

7325189087
5
it
your
reality
real
our
4294967296
5
it
your
reality
real
our
213456
12
id
a
h
lm
aid
aidhlm
hl
m
dhlm
ai
aid
hlm
2017697240238123468420176972402381234684201769724023812346842017697240238123468420176972402381234684
7
azipmwpbgqadtjafhmth
azipmwpbgqadtjafhmth
azipmwpbgqadtjafhmth
azipmwpbgqadtjafhmth
azipmwpbgqadtja
azipmwpbgqadtjafhmthazipmwpbgqadtjafhmthazipmwpbgqadtjafhmthazipmwpbgqadtjafhmth
fhmth
27386428376
0
97854668689678768
3
dfgdfhjghjdggzash
dfgkjt
nkhhdfsge
11111111111
5
ij
jj
ji
j
i
1234567890
5
bdh
txo
bdhkmp
i
kmp
-1

This is my code:

import time
import sys
sys.setrecursionlimit(100000)

n = int(input("Enter N (for digits): "))
k = int(input("Enter K: "))

rangeVar = pow(k, n) - pow(k, n-1)
starter = pow(k, n) - rangeVar
counter = 0

start = time.clock()
def calc(i):
    global starter, rangeVar, counter, n, k

    temp = str(i)

    if i == pow(k, n):
        return counter
    else:
        for j in range(len(temp) - 1):
            if temp[j] == '0' and temp[j+1] == '0':
                counter += 1
        calc(i+1)

calc(starter)
end = time.clock()
print(rangeVar - counter)
print(end - start)

Remove repetitions before applying your algo i.e.: sequences like 2 1 2 2 2 2 1 change into 2 1 2 1
It helped me with WA#21

This test is something like this
2 2
1 2
1 2
1 2
Correct answer is YES

I am get WA on test 4.

Need help. Stuck for serval weeks!
What is wa3 like?
What is wa6 like?

Store a sorted array [L...R]  in each vertex of your ST,  corresponding to [L;R].

Скажите, пожалуйста, как можно на Вашем сайте создать свою "олимпиаду", турнир по программированию.

Please Help and give me some tests!

Edited by author 01.11.2017 20:58

If in the line "outputon", this does not mean that you need to put "out" and "puton", since sometimes you can put "output" and "one".

Сheck out this test:
1
outputone
Answer : "YES"

Perhaps this will help you.

Edited by author 01.11.2017 18:28

Edited by author 01.11.2017 18:29

 If start from all the leaf airport that from one airport could win, start from this airport suerly will lose. Otherwise,he could win.

My program had such mistake: it has printed results in the order in which athletes have finished the race, but the task is to print results in order in which athletes have started one.

Here is my code which has been written in java. Please help me. I cannot understand what wrong is. Or some tests you have ?. Thanks.

public class EsreveRredrO_1226 {
    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        String s = reader.readLine();
        solve(s);
    }

    public static void solve(String s) {
        String text[];
        text = s.split(" ");
        for (int i = 0; i < text.length; i++) {
            String revers = "";
            String temp;
            for (int j = text[i].length() - 1; j >= 0;  j--) {
                    revers = revers.concat(String.valueOf(text[i].charAt(j)));
            }
            if (revers.charAt(0) == '!' || revers.charAt(0) == ',' || revers.charAt(0) == '.' || revers.charAt(0) == '?'
                    || revers.charAt(0) == ':' || revers.charAt(0) == ' ') {
                temp = revers.substring(1, revers.length());
                revers = temp.concat(String.valueOf(revers.charAt(0)));
            }
            if (i != text.length - 1)
            System.out.print(revers + " ");
            else System.out.print(revers);
        }
    }
}

Edited by author 04.09.2017 00:45

6
1 3 2 6 5 4

answer: 0

Edited by author 31.10.2017 02:47

it's one of the most awful descriptions of the problems i've ever seen. some corner cases, which are really questionable, are not explained. it really irritates because instead of solving the next problem you need to guess what the answer is. how should i know what is the answer for 1? i am not linguist, but smth tells me, that there are no situation when you can find elements in between in the list of one element. what is the answer? should i put 0, 1, etc? after all it's not a competition, i can't even ask the judge or stuff how should i interpret the problem itself.

Edited by author 31.10.2017 02:27