CodeForces - 834A

CodeForces - 834A

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

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

Output

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

Example

Input

^ > 1

Output

cw

Input

< ^ 3

Output

ccw

Input

^ v 6

Output

undefined

题目大概：

旋转图形，直到旋转前后的样子，和旋转的次数（每次旋转90度），问是顺时针，逆时针，其他。

思路：

如果顺时针，可以转变位数字，加上旋转次数，对4除余。

逆时针，转变为数字，也是列出公式（自己推一下），对4除余。

代码：

#include using namespace std;int f(char a){if(a==118)return 1;else if(a==60)return 2;else if(a==94)return 3;else if(a==62)return 4;return 0;}int main(){char n,m;long long q1,w1,q2,w2,k;cin>>n>>m>>k;q1=q2=f(n);w1=w2=f(m);int p=0;if(k==0){if(q1==w1)p=1;}else{q1=(q1+k)%4;if(q1==0)q1=4;if(q1==w1)p=1;}int o=0;if(k==0){if(q2==w2)o=1;}else{ if(k

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