zoukankan      html  css  js  c++  java
  • Codeforces Round #260 (Div. 2) B. Fedya and Maths

    B. Fedya and Maths
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Fedya studies in a gymnasium. Fedya's maths hometask is to calculate the following expression:

    (1n + 2n + 3n + 4nmod 5

    for given value of n. Fedya managed to complete the task. Can you? Note that given number n can be extremely large (e.g. it can exceed any integer type of your programming language).

    Input

    The single line contains a single integer n (0 ≤ n ≤ 10105). The number doesn't contain any leading zeroes.

    Output

    Print the value of the expression without leading zeros.

    Sample test(s)
    Input
    4
    Output
    4
    Input
    124356983594583453458888889
    Output
    0
    Note

    Operation x mod y means taking remainder after division x by y.

    Note to the first sample:

    #include <cstdio>
    #include <cstring>
    using namespace std;
    char str[100000+10];//注意拿数组进行储存
    int main()
    {
        scanf("%s",str);
        int len = strlen(str);
        int sum = (str[len-2]-'0')*10+str[len-1]-'0';
        if(sum%4==0)puts("4");//做题前先打表
        else puts("0");
        return 0;
    }
  • 相关阅读:
    使用git bash提交代码到github托管
    电子邮件的正则表达式
    PHP正则表达式及实例
    php中session_start()函数的作用
    mysql 中文乱码
    mysql 安装以及配置
    高质量JAVA代码编写规范
    DAO设计模式
    深入浅出UML类图
    分析业务模型-类图(Class Diagram)
  • 原文地址:https://www.cnblogs.com/qscqesze/p/3900657.html
Copyright © 2011-2022 走看看