C. Longest Regular Bracket Sequence

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/problemset/problem/5/C

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

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

Sample Output

6 2

HINT

题意

给你一个括号序列,让你找到最长的连续的合法括号序列

然后让你输出这个括号序列的长度是多少

这么长的括号序列一共有多少个

题解:

看到括号匹配,就用stack来弄就好了

然后我们再简单dp一下,表示以这个字符结尾的序列的长度是多少

然后跑一发就好了

代码

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 2000001
#define mod 10007
#define eps 1e-9
int Num;
char CH[];
//const int inf=0x7fffffff; //нчоч╢С
const int inf=0x3f3f3f3f;
/* inline void P(int x)
{
Num=0;if(!x){putchar('0');puts("");return;}
while(x>0)CH[++Num]=x%10,x/=10;
while(Num)putchar(CH[Num--]+48);
puts("");
}
*/
inline ll read()
{
int x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
inline void P(int x)
{
Num=;if(!x){putchar('');puts("");return;}
while(x>)CH[++Num]=x%,x/=;
while(Num)putchar(CH[Num--]+);
puts("");
}
//************************************************************************************** string s;
stack<int> k;
int dp[maxn];
int main()
{
cin>>s;
for(int i=;i<s.size();i++)
{
if(s[i]=='(')
k.push(i);
else
{
if(!k.empty())
{
dp[i]=i-k.top()+;
if(k.top()>)
dp[i]+=dp[k.top()-];
k.pop();
}
}
}
int ans1=,ans2=;
for(int i=;i<s.size();i++)
{
if(dp[i]>ans1)
{
ans1=dp[i];
ans2=;
}
else if(dp[i]==ans1)
ans2++;
}
if(ans1==)
printf("0 1\n");
else
cout<<ans1<<" "<<ans2<<endl;
}

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