Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

Description

Stringld(left delete) is a function that gets a string and deletes its leftmost character (for instance Stringld(``acm") returns ``cm").

You are given a list of distinct words, and at each step, we apply stringld
on every word in the list. Write a program that determines the number
of steps that can be applied until at least one of the conditions become
true:

1. A word becomes empty string, or
2. a duplicate word is generated.

For example, having the list of words aab, abac, and caac, applying the function on the input for the first time results in ab, bac, and aac. For the second time, we get b, ac, and ac. Since in the second step, we have two ac
strings, the condition 2 is true, and the output of your program should
be 1. Note that we do not count the last step that has resulted in
duplicate string. More examples are found in the sample input and output
section.

Input

Output

Sample Input

4

aaba

aaca

baabcd

dcba

3

aaa

bbbb

ccccc

0

#include <cstdio>

#include <cmath>

#include <cstring>

#include <ctime>

#include <iostream>

#include <algorithm>

#include <set>

#include <vector>

#include <sstream>

#include <queue>

#include <typeinfo>

#include <fstream>

typedef long long ll;

using namespace std;

//freopen("D.in","r",stdin);

//freopen("D.out","w",stdout);

#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)

#define maxn 100001

const int inf=0x7fffffff;   //无限大

//string &erase(int pos = 0, int n = npos);//删除pos开始的n个字符，返回修改后的字符串

string s[];

int main()

{

    int n;

    while(cin>>n)

    {

        if(n==)

            break;

        for(int i=;i<n;i++)

            cin>>s[i];

        int flag=;

        int flag2=;

        if(n!=)

        while()

        {

            for(int i=;i<n;i++)

            {

                for(int j=;j<n;j++)

                {

                    if(i==j)

                        continue;

                    for(int k=;k<s[i].size();k++)

                    {

                        if(s[i][k]!='(')

                            break;

                        if(k==s[i].size()-)

                        {

                            flag2=;

                            break;

                        }

                    }

                    if(s[i]==s[j])

                    {

                        flag2=;

                        break;

                    }

                }

            }

            if(flag2==)

                break;

            for(int i=;i<n;i++)

            {

                s[i][flag]='(';

            }

            flag++;

        }

        flag--;

        if(flag<)

            flag=;

        cout<<flag<<endl;

    }

    return ;

}