You can build binary tree
Left child corresponds to inward fold
Right child corresponds to outward fold
Root is unfolded strip
Build tree from root to leaves and before going deeper check winning conditions

How to deal with case like this
' ' like ' like ' like ' like ' 'like '
it can be splitted as
( ) & ( like ' like ' like ' 'like ) or
( ' like ) & ( like ' like ' 'like ) or
( ' like ' like ) & ( like ' 'like ) or
( ' like ' like ' like ) & ( 'like )

Can you clarify this statement
Inner entrance of apostrophe symbol (ASCII 39) into string or template is encoded by double apostrophe symbol

And if it's true the case
''hi'' like ''hi''
will be changed to
'"hi'' like '"hi'
Am I right?

Then how is this case possible?
'''''' like '_'''

Does anyone know about test number 8?. If you do, please upload some, thank you very much.

#include <stdio.h>
int main()
{
    int a,b;
    scanf("%d,%d",&a,&b);
    printf("a+b=%d\n",a+b);
    return 0;
}

please, how can I get test cases on this oj?

Error in jury solution was fixed, all test answers updated, few new tests added. Most authors lost their AC.
Big thanks to the Winger for pointing out an issue and helping with fixing it.

#include <bits/stdc++.h>

using namespace std;

int main()
{
    int a, a2, a3;
    int b, b2, b3;
    scanf("%d %d %d %d %d %d", &a, &b, &a2, &b2, &a3, &b3);
    int bb1 = a - a3;
    int bb2 = b - b2;
    printf("%d %d\n", bb1, bb2);

    return 0;
}

#include<stdio.h>
#include<math.h>
#include<stdlib.h>
#define SIZE 128*1024
int main()
{
    int i,n=0;
    long long int num[SIZE],N;

printf("enter your aray value:\n");
    while(scanf("%lld",&N)!=EOF)
          {
              num[n]=N;
            n++;
        }

    for(i=n-1;i>=0;i--){

      printf("%0.4lf\n",(double)sqrt(num[i]));
    }
    return 0;
}

4th test is:

3
0 0
9999 10000
9998 9999

from re import *
from math import sqrt

result = findall('[0-9]+', input())
result.reverse()

for i in result:
    print('{:.4f}'.format(sqrt(int(i))))

Edited by author 14.03.2017 01:20

Who had a Wrong Answer in 5 th test ? What mistake may be there ?

Could someone provide tests for me? Thank you

Try this test:
4 3
####
#..#
####
Answer is 1 :)

#include <bits/stdc++.h>

using namespace std;

int main()
{int x;
     vector <string> myvector (2001);
  fill (myvector.begin(),myvector.begin()+5,"few");
  fill (myvector.begin()+6,myvector.begin()+10,"several");
  fill (myvector.begin()+11,myvector.begin()+20,"pack");
  fill (myvector.begin()+21,myvector.begin()+50,"lots");
  fill (myvector.begin()+51,myvector.begin()+100,"horde");
  fill (myvector.begin()+101,myvector.begin()+250,"throng");
  fill (myvector.begin()+251,myvector.begin()+500,"swarm");
 fill (myvector.begin()+501,myvector.begin()+1000,"zounds");
 fill (myvector.begin()+1001,myvector.end(),"legion");
 cin>>x;
 cout<<myvector[x];

    return 0;
}
correct solution:
#include <bits/stdc++.h>

using namespace std;

int main()
{int x;
     vector <string> myvector (2001);
  fill (myvector.begin(),myvector.begin()+5,"few");
  fill (myvector.begin()+5,myvector.begin()+10,"several");
  fill (myvector.begin()+10,myvector.begin()+20,"pack");
  fill (myvector.begin()+20,myvector.begin()+50,"lots");
  fill (myvector.begin()+50,myvector.begin()+100,"horde");
  fill (myvector.begin()+100,myvector.begin()+250,"throng");
  fill (myvector.begin()+250,myvector.begin()+500,"swarm");
 fill (myvector.begin()+500,myvector.begin()+1000,"zounds");
 fill (myvector.begin()+1000,myvector.end(),"legion");
 cin>>x;
 cout<<myvector[x]<<endl;;

    return 0;
}

Edited by author 30.09.2017 07:20

I want to know how to approach this question in what direction to think so that i can solve it on my own.
Thank you.

Маяки могут находится в 1 точке.
P.S.
 Если считать, что маяки не могут быть в одной точке, то тогда нужно решать паросочетанием (как я в начале и решал), что гораздо сложнее по реализации

I tried different tasks and always got the right answer.

any test?