codeforces 5C Longest Regular Bracket Sequence（dp+技巧）【最长连续括号模板】

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".

Examples

input

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

output

6 2

input

))(

output

0 1

 【题解】 先说一下，这本来是个区间dp，，但是那个推公式很麻烦，所以就用了这个方法，不过原理是差不多的，

 分析： 

 题目要求一个乱序括号序列的最长连续合法序列，及此长度的合法序列个数，那么我们可以扫两次。

先从左往右扫，遇见（就num++，遇见）并且num>0，就把此）标记为1，并且num--，意味着有一对已经匹配了，剩下num--个（了。

然后同样从右往左扫，操作相同；

扫完后再次从左往右扫一遍，统计连续标记的最大值，并记录其个数即可。

【AC代码】

#include
#include
#include
#include
using namespace std;
const int N=1e6+10;
int maxn,m,n,len;
char s[N];
int vis[N];

int main()
{
    while(~scanf("%s",s))
    {
        memset(vis,0,sizeof(vis));
        maxn=-1;
        len=strlen(s);
        int cnt=0;
        for(int i=0;i=0;--i)
        {
            if(s[i]==')')
                cnt++;
            else if(s[i]=='(')
            {
                if(cnt)
                   vis[i]=1;
                cnt--;//注意--
            }
            if(cnt<0) cnt=0;//这里注意，，WA了好几发呢
        }
        cnt=0;
        n=0;
        for(int i=0;imaxn)
            {
                maxn=cnt;
                n=1;
            }
            else if(cnt == maxn)
                n++;
        }
        if(maxn==0) n=1;//特判  注意别忘了
        printf("%d %d\n",maxn,n);
    }
    return 0;
}