The Useless Toy

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本文介绍了一个有趣的编程问题：通过玩具的起始和结束位置及旋转时间推断其旋转方向。通过对不同情况的枚举和模运算，实现了对旋转方向的有效判断。

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                                    A. The Useless Toy
                                time limit per test1 second
                                memory limit per test256 megabytes
                                    inputstandard input
                                    outputstandard output

Walking through the streets of Marshmallow City, Slastyona have spotted some merchants selling a kind of useless toy which is very popular nowadays – caramel spinner! Wanting to join the craze, she has immediately bought the strange contraption.

Spinners in Sweetland have the form of V-shaped pieces of caramel. Each spinner can, well, spin around an invisible magic axis. At a specific point in time, a spinner can take 4 positions shown below (each one rotated 90 degrees relative to the previous, with the fourth one followed by the first one):

After the spinner was spun, it starts its rotation, which is described by a following algorithm: the spinner maintains its position for a second then majestically switches to the next position in clockwise or counter-clockwise order, depending on the direction the spinner was spun in.

Slastyona managed to have spinner rotating for exactly n seconds. Being fascinated by elegance of the process, she completely forgot the direction the spinner was spun in! Lucky for her, she managed to recall the starting position, and wants to deduct the direction given the information she knows. Help her do this.

Input
There are two characters in the first string – the starting and the ending position of a spinner. The position is encoded with one of the following characters: v (ASCII code 118, lowercase v), < (ASCII code 60), ^ (ASCII code 94) or > (ASCII code 62) (see the picture above for reference). Characters are separated by a single space.

In the second strings, a single number n is given (0 ≤ n ≤ 109) – the duration of the rotation.

It is guaranteed that the ending position of a spinner is a result of a n second spin in any of the directions, assuming the given starting position.

Output
Output cw, if the direction is clockwise, ccw – if counter-clockwise, and undefined otherwise.

Examples
input
^ >
1
output
cw
input
< ^
3
output
ccw
input
^ v
6
output
undefined

这是一道思维题，问<>^v是怎么旋转的，先把偶数倍的都去掉%一下，然后分开枚举用4来%，^>和>^类似的顺序相反，只有两种顺序ccw和cw

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
    int n;
    char c1,c2;
    cin>>c1>>c2;
    cin>>n;
    if(n%2==0)
        cout<<"undefined"<<endl;
    if(c1=='^')
    {
        if(c2=='>')
        {
            if(n%4==1)
                cout<<"cw"<<endl;
            else
                cout<<"ccw"<<endl;
        }
        if(c2=='<')
        {
            if(n%4==1)
                cout<<"ccw"<<endl;
            else
                cout<<"cw"<<endl;
        }
    }
    if(c1=='v')
    {
        if(c2=='>')
        {
            if(n%4==1)
                cout<<"ccw"<<endl;
            else
                cout<<"cw"<<endl;
        }
        if(c2=='<')
        {
            if(n%4==1)
                cout<<"cw"<<endl;
            else
                cout<<"ccw"<<endl;
        }
    }
    if(c1=='<')
    {
        if(c2=='^')
        {
            if(n%4==1)
                cout<<"cw"<<endl;
            else
                cout<<"ccw"<<endl;
        }
        if(c2=='v')
        {
            if(n%4==1)
                cout<<"ccw"<<endl;
            else
                cout<<"cw"<<endl;
        }
    }
    if(c1=='>')
    {
        if(c2=='^')
        {
            if(n%4==1)
            {
                cout<<"ccw"<<endl;
            }
            else
                cout<<"cw"<<endl;
        }
        if(c2=='v')
        {
            if(n%4==1)
                cout<<"cw"<<endl;
            else
                cout<<"ccw"<<endl;
        }
    }
    return 0;
}