【44.10%】【codeforces 723B】Text Document Analysis

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard output

Modern text editors usually show some information regarding the document being edited. For example, the number of words, the number of pages, or the number of characters.

In this problem you should implement the similar functionality.

You are given a string which only consists of:

uppercase and lowercase English letters,

underscore symbols (they are used as separators),

parentheses (both opening and closing).

It is guaranteed that each opening parenthesis has a succeeding closing parenthesis. Similarly, each closing parentheses has a preceding opening parentheses matching it. For each pair of matching parentheses there are no other parenthesis between them. In other words, each parenthesis in the string belongs to a matching “opening-closing” pair, and such pairs can’t be nested.

For example, the following string is valid: “Hello_Vasya(and_Petya)__bye(and_OK)”.

Word is a maximal sequence of consecutive letters, i.e. such sequence that the first character to the left and the first character to the right of it is an underscore, a parenthesis, or it just does not exist. For example, the string above consists of seven words: “Hello”, “Vasya”, “and”, “Petya”, “bye”, “and” and “OK”. Write a program that finds:

the length of the longest word outside the parentheses (print 0, if there is no word outside the parentheses),

the number of words inside the parentheses (print 0, if there is no word inside the parentheses).

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 255) — the length of the given string. The second line contains the string consisting of only lowercase and uppercase English letters, parentheses and underscore symbols.

Output

Print two space-separated integers:

the length of the longest word outside the parentheses (print 0, if there is no word outside the parentheses),

the number of words inside the parentheses (print 0, if there is no word inside the parentheses).

Examples

input

37

Hello_Vasya(and_Petya)__bye(and_OK)

output

5 4

input

37

a(b___c)de_f(g)__h__i(j_k_l)m

output

2 6

input

27

(LoooonG)shOrt(LoooonG)

output

5 2

input

5

(_)

output

0 0

Note

In the first sample, the words “Hello”, “Vasya” and “bye” are outside any of the parentheses, and the words “and”, “Petya”, “and” and “OK” are inside. Note, that the word “and” is given twice and you should count it twice in the answer.

【题解】



字符串处理。

要求括号内的单词个数以及括号外最长的单词的长度。

把括号里面的内容提取出来然后用一个”_”代替这个括号和里面的内容；

ere_(dfsfd)df把括号去掉后就变成

ere__df

然后提取出来的是

dfsfd

然后在后面加个_

即

dfsfd;

重复上述操作直到

s存的是括号外的东西

ins存的是括号内的东西；

加上_的目的是区分两个单词;

然后进行所需的操作即可

#include <string>
#include <algorithm>
#include <iostream>

using namespace std;

int n;
string s;
string ins = "";

bool is_zm(char x)
{
    if (('a' <= x && x <= 'z') || ('A' <= x && x <= 'Z'))
        return true;
    return false;
}

int main()
{
    //freopen("F:\\rush.txt", "r", stdin);
    scanf("%d", &n);
    cin >> s;
    while (true)
    {
        bool flag = false;
        int len = s.size();
        for (int i = 0; i <= len - 1; i++)
            if (s[i] == '(')
            {
                flag = true;
                int j = i;
                while (s[j] != ')')
                    j++;
                int len1 = j - i + 1;
                string temp = s.substr(i + 1, len1 - 2);
                s.erase(i, len1);
                s.insert(i, "_");
                ins += temp;
                ins += '_';
                break;
            }
        if (!flag)
            break;
    }
    s = "_" + s + "_";
    ins = "_" + ins;
    int ma = 0, cntin = 0;
    int len = s.size();
    int i = 0, j;
    while (i <= len - 1)
    {
        j = i;
        if (is_zm(s[i]))
        {
            while (is_zm(s[j + 1]))
                j++;
            int len = j - i + 1;
            ma = max(len, ma);
        }
        i = j + 1;
    }
    len = ins.size();
    i = 0;
    while (i <= len - 1)
    {
        j = i;
        if (is_zm(ins[i]))
        {
            cntin++;
            while (is_zm(ins[j + 1]))
                j++;
        }
        i = j + 1;
    }
    cout << ma << " " << cntin << endl;
    return 0;
}