CodeForces 723B-Text Document Analysis（模拟）

B. 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.

题意：给定一组由大小写字母，下划线，小括号组成的字符串，其中下划线可以看作是空格。求出括号内单词的数量和括号外最长单词的长度。

代码

#include<stdio.h>#include<iostream>#include<algorithm>#include<string.h>using namespace std;int main(){    int N;    char first,next;    scanf("%d",&N);    getchar();    scanf("%c",&first);    int num_1=0;//括号层数    int num_2=0;//括号外最长连续单词长度    int num_3=0;//当前单词长度    int num_4=0;//括号内单词总数    if(first=='(')        num_1++;    else if((first>='a'&&first<='z')||(first>='A'&&first<='Z'))        num_3++;    num_2=max(num_2,num_3);    while(N--)    {        scanf("%c",&next);        if((next>='a'&&next<='z')||(next>='A'&&next<='Z'))        {            if(num_1==0)            {                num_3++;                num_2=max(num_2,num_3);            }            else if(!((first>='a'&&first<='z')||(first>='A'&&first<='Z')))                num_4++;        }        else if(next=='(')        {            num_1++;            num_3=0;        }        else if(next==')')            num_1--;        else if(next=='_'&&num_1==0)        {            num_2=max(num_2,num_3);            num_3=0;        }        first=next;    }    printf("%d %d\n",num_2,num_4);    return 0;}