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

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

笨办法

#include<bits/stdc++.h>
using namespace std;
char sta;
char en;
long minu;
char pos[4]={'^','>','v','<'};
int main()
{
cin>>sta>>en>>minu;
int p1=0,p2=0;
for(int i=0;i<4;i++)
{
if(pos[i]==sta)
p1=i;
if(pos[i]==en)
p2=i;

}
long step=minu%4;
//int step2=3-max(p1,p2)+min(p1,p2);
//cout<<step;
if(step!=0&&step!=2)
{
if(pos[p1+step]==pos[p2])
cout<<"cw"<<endl;
else if(p1+step>3&&pos[p2]==pos[-1+step-(3-p1)])
cout<<"cw"<<endl;
else if(pos[p1-step]==pos[p2])
cout<<"ccw"<<endl;
else if(p1-step<0&&pos[p1]==pos[-1+step-(3-p2)])
cout<<"ccw"<<endl;
else
cout<<"undefined"<<endl;
}
else
cout<<"undefined"<<endl;

}

公式

#include<bits/stdc++.h>
using namespace std;
char sta;
char en;
long minu;
char pos[4]={'^','>','v','<'};
int main()
{
cin>>sta>>en>>minu;
int p1=0,p2=0;
for(int i=0;i<4;i++)
{
if(pos[i]==sta)
p1=i;
if(pos[i]==en)
p2=i;

}
long step=minu%4;
//int step2=3-max(p1,p2)+min(p1,p2);
//cout<<step;
if(step!=0&&step!=2)
{
if(pos[(p1+step)%4]==pos[p2])
cout<<"cw"<<endl;

else if(pos[abs(p2+step)%4]==pos[p1])
cout<<"ccw"<<endl;

else
cout<<"undefined"<<endl;
}
else
cout<<"undefined"<<endl;

}

可以知道，一个数组成环状时，有一个元素ak的下n个元素为a（k+n）%4,m为元素个数，同时，此元素的下k个等于彼元素的上k个，所以不需要上k个元素公式