E. String Multiplication

Roman and Denis are on the trip to the programming competition. Since the trip was long, they soon got bored, and hence decided to came up with something. Roman invented a pizza's recipe, while Denis invented a string multiplication. According to Denis, the result of multiplication (product) of strings

Roman was jealous of Denis, since he invented such a cool operation, and hence decided to invent something string-related too. Since Roman is beauty-lover, he decided to define the beauty of the string as the length of the longest substring, consisting of only one letter. For example, the beauty of the string "xayyaaabca" is equal to

In order to entertain Roman, Denis wrote down

Input

The first line contains a single integer

It's guaranteed, that the total length of the strings

Output

Print exactly one integer — the beauty of the product of the strings.

Examples

input

Copy

3
a
b
a

output

Copy

input

Copy

2
bnn
a

output

Copy

Note

In the first example, the product of strings is equal to "abaaaba".

In the second example, the product of strings is equal to "abanana".

考虑每一个字符，分情况模拟即可：
对于 $a + b$

如果 $b$

我上来是暴力计算，结果超时，gg

#include <bits/stdc++.h>
using namespace std;

#define st first
#define nd second
typedef long long LL;
typedef long double LLD;
typedef pair <int, int> PII;
typedef pair <LL, LL> PLL;
typedef pair <LLD, LLD> PLLD;

struct Info 
{
    int L, R;
    char Lc, Rc;

    Info (char Lc_, char Rc_) : L(1), R(1), Lc(Lc_), Rc(Rc_) {}
};

const int C = 26;
int D[C];

int GetInt (char c)
{
    return (int)(c - 'a');
}

void Add (string s, int T[])
{
    int cnt = 1;
    char c = s[0];

    for (int i = 1; i < s.size(); ++i)
    {
        if (s[i] == s[i - 1])
            ++cnt;
        else
        {
            cnt = 1;
            c = s[i];
        }

        int l = GetInt(c);
        T[l] = max(T[l], cnt);
    }
}

Info GetInfo (string s)
{   
    Info info = Info(s[0], s[s.size() - 1]);

    int i = 1;
    while (i < s.size() && s[i] == s[i - 1])
    {
        ++info.L;
        ++i;
    }

    i = s.size() - 2;
    while (i >= 0 && s[i] == s[i + 1])
    {
        ++info.R;
        --i;
    }

    if (info.R == s.size()) 
        info.R = 0;

    return info;
}

void Update (string s)
{
    int T[C] = {};
    for (int i = 0; i < C; ++i)
        if (D[i])
            T[i] = 1;

    Add(s, T);

    Info info = GetInfo(s);
    
    if (info.R == 0)
    {
        int l = GetInt(info.Lc), d = D[l];
        T[l] = d + (d + 1) * info.L;
    }
    else
    {
        int lL = GetInt(info.Lc), lR = GetInt(info.Rc);
        T[lL] = info.L;
        T[lR] = max(T[lR], info.R);

        if (lL == lR)
        {
            if (D[lL])
                T[lL] = info.L + info.R + 1;
        }
        else
        {
            if (D[lL])
                T[lL] = max(T[lL], info.L + 1);
            if (D[lR])
                T[lR] = max(T[lR], info.R + 1);
        }
    }

    copy(T, T + C, D);
}

int main ()
{
    ios_base::sync_with_stdio(false);

    int n;
    cin >> n;

    for (int i = 0; i < n; ++i)
    {
        string s;
        cin >> s;
        Update(s);        
    }

    int result = 0;
    for (int i = 0; i < C; ++i)
        result = max(result, D[i]);
    printf("%d", result);

    return 0;
}

上面是大佬代码

E. String Multiplication

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