括号匹配 CodeForces 5C

Description

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 Input

Input

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

Output

6 2

Input

))(

Output

0 1

#include <cstdio>
#include <queue>
#include <stack>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int oo = 1e9;
const double PI = acos(-1);
const int N = 1e6+7;
char str[N];
int len[N];/**< 如果这个位置是‘)‘ 保存从字符串头部开始到这个位置的最大匹配数量 */
int main()
{
    int i, ans=0, sum=0;
    stack <int >sta;
    scanf("%s", str);
    int M = strlen(str);
    for(i = 0; i < M; i++)
    {
        if(str[i] == ‘(‘) sta.push(i);
        else if(sta.size())
        {
            int top = sta.top();

            sta.pop();

            if(str[top-1] == ‘)‘)
                len[i] = (i-top+1) + len[top-1];

            else
                len[i] = (i-top+1);

            ans = max(ans, len[i]);
        }
    }
    for(i = M-1; i >= 0; i--)
    {
        if(str[i] == ‘)‘ && len[i] && len[i] == ans)
        sum++;
    }
    if(sum == 0) sum = 1;
    printf("%d %d\n", ans, sum);
    return 0;
}