834A. 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：

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次顺时针旋转或者n此逆时针旋转都能由第一个字符变为第二个字符，那么输出“undefined”（我觉得这一点最奇怪，为啥是都能。。。）。

#include<cstdio>#include<algorithm>#include<iostream>#include<cmath>#include<iomanip>#include<cstring>#include<vector>#include<iterator>#define N 10001using namespace std;int main(){    char a,b;    int x,i,t,m;    int y[4]={118,60,94,62};    while(cin >> a >>b)    {        scanf("%d", &x);        for(i=0; i<4; i++){            if((int)a==y[i])                t=i;            if((int)b==y[i])                m=i;        }        /*如果a在数组中的位置和x的和除以4正好等于m在数组中的位置,        证明顺时针旋转x次可以得到想要的结果*/        if((t+x)%4==m&&(((t-x)%4)+4)%4==m)            printf("undefined\n");        else if((((t-x)%4)+4)%4==m) //这里注意括号的个数            printf("ccw\n");        else            printf("cw\n");    }    return 0;}

注意！！！如果用scanf就要注意要吃字符！！！用cin用不用考虑回车占字符位的问题！！！