zoukankan      html  css  js  c++  java
  • CodeForces 917A The Monster 贪心+思维

    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.

    Example

    Input
    ((?))
    Output
    4
    Input
    ??()??
    Output
    7

    Note

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

    1. "(?" which can be transformed to "()".
    2. "?)" which can be transformed to "()".
    3. "((?)" which can be transformed to "(())".
    4. "(?))" which can be transformed to "(())".

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

    1. "??" which can be transformed to "()".
    2. "()".
    3. "??()" which can be transformed to "()()".
    4. "?()?" which can be transformed to "(())".
    5. "??" which can be transformed to "()".
    6. "()??" which can be transformed to "()()".
    7. "??()??" which can be transformed to "()()()".

    可以将?变成)或(,问括号匹配数量

    #include <cstdio>  
    #include <cstring>  
    #include <algorithm>  
    #include <iostream>
    #include <vector>
    #include <string>
    #include <stack>
    #define debug(a)  cout << #a << " = "  << a <<endl
    using namespace std;
    int main() {
        string s;
        while( cin >> s ) {
            int ans = 0;
            for(int i=0;i<s.length();i++) { //枚举所有情况下?对字符串的影响
                int cnt1 = 0,cnt2 = 0; //cnt1把?看作)未匹配数量,cnt2把?看作(未匹配数量 
                for(int j=i;j<s.length();j++) {
                    if( s[j] == '(' ) {
                        cnt1 ++;
                        cnt2 ++;
                    } else if( s[j] == ')' ) {
                        cnt1 --;
                        cnt2 --;
                    } else {
                        cnt1 --;
                        cnt2 ++;
                    }
                    if( cnt1 < 0 && cnt2 > 0 ) {
                        cnt1 += 2; //当把?看成)导致cnt1<0时再把?看成(,这样未匹配数量又为1了
                        //cnt2 --;
                    } 
                    if( cnt2 < 0 ) {
                        break;
                    }
                    if( cnt1 == 0 ) {
                        ans ++;
                    }
                }
                //debug(i);
                //debug(ans);
            }
            cout << ans << endl;
        }
        return 0;
    }
    彼时当年少,莫负好时光。
  • 相关阅读:
    HTTP协议图解
    .NET 发布网站步骤
    使用php在服务器端生成图文验证码
    SQLServer复习文档1(with C#)
    理解 JavaScript 原型 / 原型链
    浅谈瀑布流
    懒加载
    jQuery ajax
    jQuery 动画效果 与 动画队列
    jQuery 事件
  • 原文地址:https://www.cnblogs.com/l609929321/p/8394706.html
Copyright © 2011-2022 走看看