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Description
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:
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 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.
Print two space-separated integers:
37
_Hello_Vasya(and_Petya)__bye_(and_OK)
5 4
37
_a_(_b___c)__de_f(g_)__h__i(j_k_l)m__
2 6
27
(LoooonG)__shOrt__(LoooonG)
5 2
5
(___)
0 0
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：求括号里面单词的个数，和括号外面单词的最长长度
解法：flag标记是否进入了 ( 还是出去了 ) ，进入 ( ,判断s[i]和s[i+1]，出去 ) ,计算单词长度，用MAX纪录最大值
#include<bits/stdc++.h>
using namespace std;
int n;
char s[300];
int main()
{
int flag=0;
int flag1=0;
int sum=0;
int num=0;
int MAX=-1;
scanf("%d%s",&n,s);
for(int i=0; i<strlen(s); i++)
{
if(s[i]==‘(‘)
{
flag1=1;
}
else if(s[i]==‘)‘)
{
flag1=0;
}
if(flag1==0)
{
// MAX=max(MAX,sum);
if((s[i]>=‘A‘&&s[i]<=‘Z‘||s[i]>=‘a‘&&s[i]<=‘z‘))
{
// cout<<s[i]<<endl;
sum++;
// cout<<sum<<endl;
MAX=max(MAX,sum);
}
else
{
//cout<<sum<<endl;
MAX=max(MAX,sum);
// cout<<MAX<<endl;
sum=0;
}
}
}
//cout<<sum<<endl;
for(int i=0;i<strlen(s);i++)
{
if(s[i]==‘(‘)
{
flag=1;
}
else if(s[i]==‘)‘)
{
flag=0;
}
if(flag==1)
{
if(s[i]>=‘A‘&&s[i]<=‘Z‘||s[i]>=‘a‘&&s[i]<=‘z‘)
{
if(s[i+1]==‘_‘||s[i+1]==‘)‘||s[i+1]==‘(‘&&i+1<strlen(s))
{
num++;
}
}
// cout<<s[i]<<endl;
}
}
cout<<MAX<<" "<<num<<endl;
return 0;
}
Codeforces Round #375 (Div. 2) B
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原文地址：http://www.cnblogs.com/yinghualuowu/p/5929816.html
