CodeForces - 918C - The Monster

C. The Monster

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

As Will is stuck in the Upside Down, he can still communicate with his mom, Joyce, through the Christmas lights (he can turn them on and off with his mind). He can't directly tell his mom where he is, because the monster that took him to the Upside Down will know and relocate him.

Thus, he came up with a puzzle to tell his mom his coordinates. His coordinates are the answer to the following problem.

A string consisting only of parentheses ('(' and ')') is called a bracket sequence. Some bracket sequence are called correct bracket sequences. More formally:

Empty string is a correct bracket sequence.
if s is a correct bracket sequence, then (s) is also a correct bracket sequence.
if s and t are correct bracket sequences, then st (concatenation of s and t) is also a correct bracket sequence.

A string consisting of parentheses and question marks ('?') is called pretty if and only if there's a way to replace each question mark with either '(' or ')' such that the resulting string is a non-empty correct bracket sequence.

Will gave his mom a string s consisting of parentheses and question marks (using Morse code through the lights) and his coordinates are the number of pairs of integers (l, r) such that 1 ≤ l ≤ r ≤ |s| and the string slsl + 1... sr is pretty, where si is i-th character of s.

Joyce doesn't know anything about bracket sequences, so she asked for your help.

Input

The first and only line of input contains string s, consisting only of characters '(', ')' and '?' (2 ≤ |s| ≤ 5000).

Output

Print the answer to Will's puzzle in the first and only line of output.

Examples

input

Copy

((?))

output

Copy

input

Copy

??()??

output

Copy

Note

For the first sample testcase, the pretty substrings of s are:

"(?" which can be transformed to "()".
"?)" which can be transformed to "()".
"((?)" which can be transformed to "(())".
"(?))" which can be transformed to "(())".

For the second sample testcase, the pretty substrings of s are:

"??" which can be transformed to "()".
"()".
"??()" which can be transformed to "()()".
"?()?" which can be transformed to "(())".
"??" which can be transformed to "()".
"()??" which can be transformed to "()()".
"??()??" which can be transformed to "()()()".

题意：给一个字符串，字符串只包含 '('，'?'，')'，1> ."" 2> . S合法 (S) 合法 3> .S,T合法，ST合法。问有多少合法的？

题解：暴力，定义两个变量 num1（代表 '(' 的数量）,num2（代表 '?' 取代 ')' 与 '(' 匹配的数量）,

当遇到 '(' 时，num1和num2需要同时增 1；

当遇到 ')' 时，num1和num2需要同时减 1;

当遇到 '?' 时，num1需要减 1，num2需要增 1; 此时默认 '?' 是 ')' ；

当遇到 ')' 时，num1和num2需要同时减 1;

当 num2 小于 0 时，意味着 ')' 的数量大于 '(' 与 '?' ，是不可能匹配的，break;

当 num1 小于 0 时，意味着 '(' 的数量小于 '(' ，需要将 '?' 转移一部分给 '(' ;

当 num1 等于 0 时，意味着匹配成功，sum++;

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<iostream>
using namespace std;
char s[5005];
int main(){
    scanf("%s",&s);
    int len=strlen(s);
    int num1,num2,sum=0;
    for(int i=0;i<len;i++){
        num1=0,num2=0;
        for(int j=i;j<len;j++){
            if(s[j]=='(')
                num1++,num2++;
            else if(s[j]==')')
                num1--,num2--;
            else
               num1--,num2++;
            if(num2<0)
                break;
            if(num1<0)
                num1+=2;  // //把?当成右括号不行了,由于之前num1--,所以num1+=2,代表把一个?当成了左括号
            if(num1==0)
                sum++;
        }
    }
    printf("%d\n",sum);
    return 0;
}

CodeForces - 918C - The Monster

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