Codeforces 5C. Longest Regular Bracket Sequence

C. Longest Regular Bracket Sequence

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

This is yet another problem dealing with regular bracket sequences.

We should remind you that a bracket sequence is called regular, if by inserting «+» and «1» into it we can get a correct mathematical expression. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not.

You are given a string of «(» and «)» characters. You are to find its longest substring that is a regular bracket sequence. You are to find the number of such substrings as well.

Input

The first line of the input file contains a non-empty string, consisting of «(» and «)» characters. Its length does not exceed 106.

Output

Print the length of the longest substring that is a regular bracket sequence, and the number of such substrings. If there are no such substrings, write the only line containing "0 1".

Sample test(s)

input

)((())))(()())

output

6 2

input

))(

output

0 1

题意：求最长的有效括号序列长度

方法： 使用一个栈辅助匹配， 每次匹配成功一个记录‘）’对应的长度

代码：

#include <stdio.h>#include <string>#include <iostream>using namespace std;const int MAXN =  1000005;int num[MAXN], leftBracket[MAXN];int main(){#ifdef _LOCAL    freopen("F://input.txt", "r", stdin);#endif  string str;  cin >> str;  int count = 0, max = 0, times = 1;  for(int i = 0; i < str.size(); ++i)  {      if(str[i] == '(')          leftBracket[++count] = i;      else if(count)      {          int pairLeft = leftBracket[count];          num[i] = i - pairLeft + 1 + num[pairLeft - 1];          --count;          if(num[i] == max)          {              ++times;          }          else if(num[i] > max)          {              max = num[i];              times = 1;          }      }  }  printf("%d %d\n", max, times);  return 0;}