zoukankan      html  css  js  c++  java
  • Case of the Zeros and Ones 分类: CF 2015-07-24 11:05 15人阅读 评论(0) 收藏

    A. Case of the Zeros and Ones
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    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 twoadjacent 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 lengthn - 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 test(s)
    Input
    4
    1100
    
    Output
    0
    
    Input
    5
    01010
    
    Output
    1
    
    Input
    8
    11101111
    
    Output
    6
    CF的一道水题
    
    #include <iostream>
    #include <cstdio>
    #include <cmath>
    #include <cstdlib>
    #include <cstring>
    #include <stack>
    #include <algorithm>
    
    using namespace std;
    
    const int MAX = 200200;
    
    char str[MAX];
    
    int main()
    {
        int n;
        scanf("%d",&n);
        scanf("%s",str);
        int a=0,b=0;
        for(int i=0;i<n;i++)
        {
            if(str[i]=='0')
            {
                a++;
            }
            else
            {
                b++;
            }
        }
        int sum=abs(a-b);
        printf("%d
    ",sum);
        return 0;
    
    }
    


    版权声明:本文为博主原创文章,未经博主允许不得转载。

  • 相关阅读:
    Spring事务(三)事务增强器
    Spring事务(二)事务自定义标签
    Spring事务(一)JDBC方式下的事务使用示例
    Spring整合MyBatis(五)MapperScannerConfigurer
    Spring整合MyBatis(四)MapperFactoryBean 的创建
    BOS物流管理系统-第五天
    BOS物流管理系统-第一天
    SSM
    【剑指offer】翻转单词顺序,C++实现
    【特征选择】嵌入式特征选择法
  • 原文地址:https://www.cnblogs.com/juechen/p/4721960.html
Copyright © 2011-2022 走看看