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  • CodeForces 556A 解题心得

    原题:

    Description

    Andrewid the Android is a galaxy-famous detective. In his free time he likes to think about strings containing zeros and ones.

    Once he thought about a string of length n consisting of zeroes and ones. Consider the following operation: we choose any two adjacent positions in the string, and if one them contains 0, and the other contains 1, then we are allowed to remove these two digits from the string, obtaining a string of length n - 2 as a result.

    Now Andreid thinks about what is the minimum length of the string that can remain after applying the described operation several times (possibly, zero)? Help him to calculate this number.

    Input

    First line of the input contains a single integer n (1 ≤ n ≤ 2·105), the length of the string that Andreid has.

    The second line contains the string of length n consisting only from zeros and ones.

    Output

    Output the minimum length of the string that may remain after applying the described operations several times.

    Sample Input

    Input
    4
    1100
    Output
    0
    Input
    5
    01010
    Output
    1
    Input
    8
    11101111
    Output
    6

    Hint

    In the first sample test it is possible to change the string like the following: .

    In the second sample test it is possible to change the string like the following: .

    In the third sample test it is possible to change the string like the following: .

     分析:我是使用栈的做法,每次判断当前栈顶的元素和待入栈的元素,如果是一对就不入栈并且弹出

            如果不是 ,就入栈

    我的代码

    #include<iostream>
    #include<cstdio>
    #include<stack>
    
    
    using namespace std;
    stack<char> s;
    
    int main()
    {
        
        int n;
        cin >> n;
        getchar();
        char ch,temp;
        while (n--)
        {
            scanf("%c", &ch);
            //cout << ch << endl;
            if (s.empty())
                s.push(ch);
            else
            {
                temp = s.top();
                if (temp == ch)        
                    s.push(ch);
                else
                {
                    s.pop();
                }
            }
        }
        cout << s.size() << endl;
    
    
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/shawn-ji/p/4696568.html
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