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".

Examples

input

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

output

6 2

input

))(

output

0 1

题目大意：给出一个括号字符串，求出它的最大合法字串的长度和最大长度的个数，当最大合法字串长度为0时，输出“0 1”。

思路：利用栈进行维护，每读入一个左括号就将它的下标压栈，每读入一个右括号就将它出栈，因为每一个合法的括号匹配，右括号一定与当前的栈顶左括号匹配，令一个vis[]数组，初始化为0，每次读到右括号就将当前位置和栈顶位置的vis置为1，最后遍历一遍vis数组，找出最大长度连续1的个数以及最大长度的个数即可。注意：最大长度为0时答案需要特判。

参考代码：

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<queue>#include<stack>#include<vector>#include<cmath>#include<set>#include<map>#include<cstdlib>#include<functional>#include<climits>#include<cctype>#include<iomanip>using namespace std;typedef long long ll;#define INF 0x3f3f3f3f#define mod 1e9+7#define clr(a,x) memset(a,x,sizeof(a))const double eps = 1e-6;stack <int> Q;char str[1000010];int vis[1000010];void Clear(){    while(!Q.empty())        Q.pop();}int main(){    while(scanf("%s",str)==1)    {        int len=strlen(str);        int ans=0;   //记录最大长度的个数        int maxn=0;  //记录最大长度        int cnt=0;   //计数器                Clear();    //每次都要将栈清空        clr(vis,0);  //将vis初始化为0        for(int i=0;i<len;i++)        {            if(str[i]=='(')                Q.push(i);            if(str[i]==')')            {                if(!Q.empty())                {                    vis[i]=1;                    vis[Q.top()]=1;     //将栈顶位置和当前位置的vis置为1                    Q.pop();                }                else                    continue;            }        }        vis[len]=0;  //强行将lenth位置的元素置为0，防止继续遍历下去        for(int i=0;i<=len;i++)        {            if(vis[i]==1)            {                cnt++;        //如果vis为1，就++            }            if(vis[i]==0)            {                if(maxn<cnt)   //读到0时，比较最大长度与读到0之前1的个数，如果小于                {                    maxn=cnt;        //每次都找到最大的长度                    ans=1;           //说明最大的长度更长，则，最大长度的数量重现置为1                    cnt=0;           //将cnt清0                }                else if(maxn==cnt)  //如果相等，就将最大长度数量加一                {                    ans++;                    cnt=0;                }                else                   //如果大于，只将cnt清零                    cnt=0;            }            }        if(maxn)            cout<<maxn<<" "<<ans<<endl;              else            puts("0 1");          //当最后遍历完之后发现maxn就每变过，说明就没有找到一个合法匹配的括号，则输出0 1    }    return 0;}