括弧匹配 CodeForces 5C

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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 10^6.

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];/**< 如果這個位置是‘)‘ 儲存從字串頭部開始到這個位置的最大匹配數量 */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;}

　　

括弧匹配 CodeForces 5C