Codeforces Round #426 (Div. 2) A. The Useless Toy

A. The Useless Toy

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard 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

 题目大意：判断第二个字符是不是第一个字符通过n步逆时针或顺时针得到，如果得不到准确的答案，输出undefined。

思路：模拟。取n%4的值，如果为0或2输出undefined，因为这个时候如果答案对，那么我们分不出顺逆。答案不对更是undefined。

这样只要处理余数为1,3 的情况了。

AC代码如下：

#include<iostream>
#include<stdio.h>
using namespace std;
char x[]={'v','<','^','>'};
char x2[]={'v','>','^','<'};
int main()
{
    char a,c;
    int n;
    cin>>a>>c;
    scanf("%d",&n);
    if(n%4==2||n%4==0){
        printf("undefined
");
    }else{
        int now,nxt=n;
        bool flag=false;
        for(int i=0;i<4;i++){
            if(x[i]==a){
                now=i;
                break;
            }
        }
        for(int i=now;;i++){
            i=i%4;
            if(i==(nxt+now)%4){
                if(x[i]==c)
                    flag=true;
                break;
            }
        }
        if(flag){
            cout<<"cw"<<endl;
            return 0;
        }else{
            for(int i=0;i<4;i++){
                if(x2[i]==a){
                    now=i;
                    break;
                }
            }
            for(int i=now;;i++){
                i=i%4;
                if(i==(nxt+now)%4){
                    if(x2[i]==c)
                        flag=true;
                    break;
                }
            }
        }
        if(flag){
            cout<<"ccw"<<endl;
            return 0;
        }else{
        cout<<"undefined"<<endl;
        return 0;
        }
    }
    return 0;
}