Codeforces723 B. Text Document Analysis （模拟/水）

题目连接：http://codeforces.com/contest/723/problem/B

B. 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).

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.

题目大意：统计括号外最长的连续字母的最大长度以及括号里有多少个连续的字母。

解题思路：直接看样例然后就直接搞。这题面真不想去看呀。

/* ***********************************************┆  ┏┓　　　┏┓ ┆┆┏┛┻━━━┛┻┓ ┆┆┃　　　　　　　┃ ┆┆┃　　　━　　　┃ ┆┆┃　┳┛　┗┳　┃ ┆┆┃　　　　　　　┃ ┆┆┃　　　┻　　　┃ ┆┆┗━┓　马　┏━┛ ┆┆　　┃　勒　┃　　┆　　　　　　┆　　┃　戈　┗━━━┓ ┆┆　　┃　壁　　　　　┣┓┆┆　　┃　的草泥马　　┏┛┆┆　　┗┓┓┏━┳┓┏┛ ┆┆　　　┃┫┫　┃┫┫ ┆┆　　　┗┻┛　┗┻┛ ┆************************************************ */#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <stack>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <bitset>using namespace std;#define rep(i,a,b) for (int i=(a),_ed=(b);i<=_ed;i++)#define per(i,a,b) for (int i=(b),_ed=(a);i>=_ed;i--)#define pb push_back#define mp make_pairconst int inf_int = 2e9;const long long inf_ll = 2e18;#define inf_add 0x3f3f3f3f#define mod 1000000007#define LL long long#define ULL unsigned long long#define MS0(X) memset((X), 0, sizeof((X)))#define SelfType intSelfType Gcd(SelfType p,SelfType q){return q==0?p:Gcd(q,p%q);}SelfType Pow(SelfType p,SelfType q){SelfType ans=1;while(q){if(q&1)ans=ans*p;p=p*p;q>>=1;}return ans;}#define Sd(X) int (X); scanf("%d", &X)#define Sdd(X, Y) int X, Y; scanf("%d%d", &X, &Y)#define Sddd(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z)#define reunique(v) v.resize(std::unique(v.begin(), v.end()) - v.begin())#define all(a) a.begin(), a.end()#define   mem(x,v)      memset(x,v,sizeof(x))typedef pair<int, int> pii;typedef pair<long long, long long> pll;typedef vector<int> vi;typedef vector<long long> vll;inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,rx=getchar();return ra*fh;}//#pragma comment(linker, "/STACK:102400000,102400000")char s[300];int main(){//freopen("in.txt","r",stdin);//freopen("out.txt","w",stdout);ios::sync_with_stdio(0);cin.tie(0);int n;n = read();scanf("%s",s);int mx = 0,sum = 0;for(int i=0;i<n;)    {        while(s[i]=='_' && i<n) i++;        int p = 1;        if(s[i]!='(')        {            int cnt = 0;            while(((s[i]>='a'&&s[i]<='z') || (s[i]>='A'&&s[i]<='Z')) && i<n)            {                i++;                cnt++;            }            mx = max(cnt,mx);        }        else        {            i++;            while(s[i]!=')')            {                while(s[i]=='_' && i<n) i++;                int flag = 0;                while(((s[i]>='a'&&s[i]<='z') || (s[i]>='A'&&s[i]<='Z')) && i<n)                {                    i++;                    flag = 1;                }                if(flag)sum++;            }            i++;        }    }printf("%d %d\n",mx,sum);return 0;}