• CodeForces 愚人节题目


    A题,1~40界美国总统名字:
    A. Mysterious strings
    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output
    Input

    The input contains a single integer a (1≤a≤40).

    Output

    Output a single string.

    Sample test(s)
    input
    2
    output
    Adams
    input
    8
    output
    Van Buren
    input
    29
    output
    Harding
    #include <iostream>

    using namespace std;

    int main()
    {
        string a[60]={"oHoH","Washington","Adams","Jefferson","Madison","Monroe","Adams"
        ,"Jackson","Van Buren","Harrison","Tyler","Polk","Taylor","Fillmore","Pierce",
        "Buchanan","Lincoln","Johnson","Grant","Hayes","Garfield","Arthur","Cleveland","Harrison",
        "Cleveland","McKinley","Roosevelt","Taft","Wilson","Harding","Coolidge","Hoover",
        "Roosevelt","Truman","Eisenhower","Kennedy","Johnson","Nixon","Ford","Carter","Reagan",
        "Bush","Clinton","Bush","Obama"};

        int n;
        cin>>n;
        cout<<a[n]<<endl;

        return 0;
    }
     


    这是B题,给你个二维码扫描一下就出来一个网址,然后就可以做了:
    B. QR code
    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    CodeForces 愚人节题目 - qhn999 - 码代码的猿猿

    Input

    The input contains two integers a1,a2 (0≤ai≤32), separated by a single space.

    Output

    Output a single integer.

    Sample test(s)
    input
    1 1
    output
    0
    input
    3 7
    output
    0
    input
    13 10
    output
    1

    #include <iostream>

    using namespace std;

    int main()
    {
        char a[33][33]={1,1,1,1,1,1,1,0,1,0,1,0,1,0,1,1,1,1,1,0,0,1,0,1,0,0,1,1,1,1,1,1,1,
                        1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,1,0,1,1,0,0,0,1,0,0,0,0,0,1,
                        1,0,1,1,1,0,1,0,0,1,1,0,1,1,0,0,0,0,0,1,1,0,1,0,0,0,1,0,1,1,1,0,1,
                        1,0,1,1,1,0,1,0,1,0,1,1,0,0,1,0,0,1,1,1,1,1,0,1,0,0,1,0,1,1,1,0,1,
                        1,0,1,1,1,0,1,0,1,1,0,0,0,1,1,0,0,0,1,1,1,1,0,0,1,0,1,0,1,1,1,0,1,
                        1,0,0,0,0,0,1,0,1,0,1,0,1,0,1,0,1,1,0,1,0,0,0,0,1,0,1,0,0,0,0,0,1,
                        1,1,1,1,1,1,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,1,1,1,1,1,1,
                        0,0,0,0,0,0,0,0,1,1,1,1,1,0,1,1,1,1,1,0,0,1,1,1,1,0,0,0,0,0,0,0,0,
                        1,0,0,0,1,0,1,1,1,1,0,0,1,0,0,0,0,1,0,1,1,1,1,0,1,1,1,1,1,1,0,0,1,
                        1,1,0,1,1,1,0,0,1,1,1,1,1,1,1,1,0,0,1,0,0,0,0,1,0,0,0,1,0,1,1,0,0,
                        0,1,1,1,0,0,1,1,1,0,1,0,0,0,0,1,0,1,0,0,0,1,1,1,0,1,0,0,0,1,0,1,0,
                        0,1,1,1,1,0,0,0,0,1,1,0,0,0,1,1,1,1,1,1,0,1,0,1,1,0,0,0,0,0,0,1,1,
                        1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,1,1,1,0,0,1,0,0,1,0,1,1,0,0,0,
                        1,1,1,0,0,0,0,1,0,1,1,1,0,1,0,0,1,1,0,1,0,0,1,1,0,1,0,1,0,0,1,0,0,
                        1,0,1,0,1,0,1,0,0,0,1,0,1,1,0,0,1,0,1,1,0,1,0,1,0,1,0,0,0,0,0,1,0,
                        1,0,1,1,0,0,0,0,0,1,0,1,0,1,0,0,0,1,1,1,1,1,0,1,0,0,0,0,0,0,0,0,0,
                        0,0,0,0,1,0,1,0,0,0,1,1,0,0,1,1,0,1,0,0,0,1,1,1,1,0,1,0,1,1,0,1,0,
                        1,0,1,0,0,1,0,0,1,1,1,1,1,0,1,1,1,1,0,0,0,1,0,1,0,1,0,0,0,1,1,1,0,
                        1,0,1,1,0,1,1,1,1,1,1,1,0,0,0,1,0,0,1,0,0,0,0,1,1,1,0,0,0,1,0,0,0,
                        0,0,0,0,1,0,0,1,1,0,0,0,1,0,0,1,1,0,0,0,0,0,1,1,0,1,0,0,0,0,0,1,0,
                        0,0,1,1,0,1,1,0,1,0,0,1,1,0,1,1,1,0,0,1,0,0,1,0,0,1,1,0,1,1,0,0,0,
                        0,1,1,1,0,1,0,1,1,0,1,0,0,0,1,0,0,0,1,1,1,1,0,1,0,1,0,1,0,0,1,1,0,
                        1,1,1,0,1,0,1,0,0,1,1,0,0,1,1,1,0,1,0,0,1,1,0,1,0,0,0,0,0,1,1,1,0,
                        1,1,0,0,0,1,0,1,0,0,1,0,1,0,1,1,1,1,0,0,0,1,0,1,1,1,1,1,1,1,0,0,0,
                        0,0,1,0,0,0,1,1,1,0,1,1,1,0,0,0,0,1,0,1,0,1,1,0,1,1,1,1,1,0,0,0,0,
                        0,0,0,0,0,0,0,0,1,1,1,0,0,1,0,1,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,1,0,
                        1,1,1,1,1,1,1,0,1,0,0,0,1,0,1,1,1,1,0,0,0,1,1,0,1,0,1,0,1,1,0,1,0,
                        1,0,0,0,0,0,1,0,0,1,1,1,0,1,0,1,0,1,1,1,1,1,0,0,1,0,0,0,1,1,0,1,1,
                        1,0,1,1,1,0,1,0,1,0,0,1,0,1,0,0,0,0,1,0,1,0,0,0,1,1,1,1,1,1,0,0,0,
                        1,0,1,1,1,0,1,0,0,0,1,1,0,1,0,0,1,0,0,1,0,1,1,1,1,1,1,0,1,1,0,1,0,
                        1,0,1,1,1,0,1,0,0,1,0,0,0,1,1,0,1,1,1,1,0,1,1,0,1,0,1,1,1,0,0,0,0,
                        1,0,0,0,0,0,1,0,0,1,1,0,0,1,1,0,0,1,1,1,1,1,0,0,1,1,1,1,0,0,0,0,0,
                        1,1,1,1,1,1,1,0,1,1,0,1,0,0,0,1,0,1,0,0,1,1,0,1,1,1,0,0,1,0,0,0,1};

        int x,y;

        cin>>x>>y;
        cout<<(int)a[x][y];

        return 0;
    }
  • 相关阅读:
    static 静态
    纽扣电池带负载能力差
    JAVA--异常(1)
    【DP专题】——洛谷P1273有线电视网
    我到现在都没有搞明白git233333
    git常见问题之git pull origin master时fatal: refusing to merge unrelated histories
    矩阵内积转化为求矩阵乘积的迹
    矩阵分解系列三:非负矩阵分解及Python实现
    矩阵分解系列三:可对角化矩阵的谱分解
    矩阵分解系列二:正交三角分解(UQ、QR分解)
  • 原文地址:https://www.cnblogs.com/CKboss/p/3351116.html
Copyright © 2020-2023  润新知