The Useless Toy

题目链接  http://codeforces.com/problemset/problem/834/A

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

^ v6

output

undefined

题目大意：

按照第一组测试数据说吧，这样明了。第一行输入两个箭头（字符），第二行输入n。如果经过n次顺时针旋转，能够由第一个字符编程第二个字符，并且经过n此逆时针旋转，不能由第一个字符变为第二个字符，那么输出“cw”；如果经过n此顺时针旋转不能由第一个字符变为第二个字符，并且经过n此逆时针旋转能够由第一个字符变为第二个字符，输出“ccw”；如果经过n此顺时针旋转或者n此逆时针旋转都不能由第一个字符变为第二个字符，那么输出“undefined”。

解题思路：

我的解题思路是笨方法，就是不断地进行判断循环，不建议看本人的代码，理解题意后，自己进行写代码。

代码：

#include<iostream>using namespace std;int main(){    char c1,c2,z1,z2,znn,zss;    int as[4]={-58,34,-32,56};    int an[4]={32,-34,58,-56};    long long int n,len;    while(cin>>c1>>c2)    {        znn=0;        zss=0;        cin>>n;        z1=c1;        z2=c2;        int zs=0;        int zn=0;        n=n%4;        len=n;        if(n==0)        {            cout<<"undefined"<<endl;            continue;        }        if(c1==118)        {            for(int i=0;;i++)            {                n--;                c1=c1+as[i];                if(n==0&&c1!=c2)                {                    zss=3;                    break;                }                if(c1==c2&&n==0)                {                    zs=1;                    break;                }            }        }        if(c1==60&&zs==0&&zss!=3)        {            for(int i=1;;i++)            {                if(i==4)                    i=0;                n--;                c1=c1+as[i];                if(n==0&&c1!=c2)                {                    zss=3;                    break;                }                if(c1==c2&&n==0)                {                    zs=1;                    break;                }            }        }        if(c1==94&&zs==0&&zss!=3)        {            for(int i=2;;i++)            {                if(i==4)                    i=0;                n--;                c1=c1+as[i];                if(n==0&&c1!=c2)                {                    zss=3;                    break;                }                if(c1==c2&&n==0)                {                    zs=1;                    break;                }            }        }        if(c1==62&&zs==0&&zss!=3)        {            for(int i=3;;i++)            {                if(i==4)                    i=0;                n--;                c1=c1+as[i];                if(n==0&&c1!=c2)                {                    zss=3;                    break;                }                if(c1==c2&&n==0)                {                    zs=1;                    break;                }            }        }        if(z1==62)        {            for(int i=0;;i++)            {                len--;                z1=z1+an[i];                if(len==0&&z1!=z2)                {                    znn=3;                    break;                }                if(z1==z2&&len==0)                {                    zn=1;                    break;                }            }        }        if(z1==94&&zn==0&&znn!=3)        {            for(int i=1;;i++)            {                if(i==4)                    i=0;                z1=z1+an[i];                len--;                if(len==0&&z1!=z2)                {                    znn=3;                    break;                }                if(z1==z2&&len==0)                {                    zn=1;                    break;                }            }        }        if(z1==60&&zn==0&&znn!=3)        {            for(int i=2;;i++)            {                if(i==4)                    i=0;                z1=z1+an[i];                len--;                if(len==0&&z1!=z2)                {                    znn=3;                    break;                }                if(z1==z2&&len==0)                {                    zn=1;                    break;                }            }        }        if(z1==118&&zn==0&&znn!=3)        {            for(int i=3;;i++)            {                if(i==4)                    i=0;                len--;                z1=z1+an[i];                if(len==0&&z1!=z2)                {                    znn=3;                    break;                }                if(z1==z2&&len==0)                {                    zn=1;                    break;                }            }        }        if(zs==1&&zn==0)            cout<<"cw"<<endl;        if(zs==1&&zn==1)            cout<<"undefined"<<endl;        if(zs==0&&zn==1)            cout<<"ccw"<<endl;    }    return 0;}