Text Document Analysis
time limit per test 1 second memory limit per test 256 megabytes
input standard input output standard 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).

传送门:CF723B

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).

Sample Input

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2
37
_Hello_Vasya(and_Petya)__bye_(and_OK)

Sample Output

1
5 4

题解

单词可以左右没有括号和下划线,注意不要理解错题意!
## AC code:(不包含输入类)
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import java.io.*;  
import java.util.*;
public class Main {

public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
String s=sc.next();
int flag1=0;
int flag2=0;
int flag3=0;
int max=-1;
int cur=0;
int cur2=0;
int number=0;
boolean start=false;
for(int i=0;i<n;i++){

if(flag2==0){
if(isAlpha(s.charAt(i))){
cur++;
//System.out.println(cur);
}
if(cur>0&&s.charAt(i)=='_'){
max=Math.max(max, cur);
cur=0;
}
if(cur>0&&s.charAt(i)=='('){
max=Math.max(max, cur);
cur=0;
}

}
if(flag2==1){
if(isAlpha(s.charAt(i))){
cur2++;
}
if(s.charAt(i)=='_'||s.charAt(i)==')'){
if(cur2>0){
number++;
cur2=0;
}
if(s.charAt(i)==')'){
flag2=0;
}
}
}
if(s.charAt(i)=='('){
flag2=1;
}


}
if(cur>0){
max=Math.max(max, cur);
cur=0;
}
if(max==-1){
System.out.print(0+" ");
}
else
System.out.print(max+" ");
System.out.println(number);
}
static boolean isAlpha(char a){
if((a>='A'&&a<='Z')||(a>='a'&&a<='z')){
return true;
}
else
return false;
}
}