题目链接：https://codeforces.com/problemset/problem/834/A
A. The Useless Toy
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard 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
旋转微调器后，它开始旋转，这由以下算法描述：微调器保持其位置一秒然后庄严地按顺时针或逆时针顺序切换到下一个位置，具体取决于旋转器旋转的方向。
题意：给一个初始位置方向和一个末位置方向，还有一个数字代表秒，每一秒旋转顺时针或者逆时针90°到下一个位置；
让你判断从初位置方向到末位置方向是顺时针还是逆时针旋转或者说两个都可以。
解题思路：一个就四个方向，我们只需要判断四个方向对应的情况就可以。
第一步：秒数取余4去掉重复旋转；
第二步：四个初始位置方向对应情况的分析判断，输出结果；
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string>
#include <string.h>
#include <queue>
#include <map>
#define MAX 100002
using namespace std;
int main()
{
char c1,c2;
while(scanf("%c %c",&c1,&c2)!=EOF)
{
int n;
scanf("%d",&n);
int m=n%4;
if(c1=='^')//上
{
if((c2=='>'&&m==1)||(c2=='<'&&m==3))//顺时针
{
printf("cw\n");
continue;
}
else if((c2=='<'&&m==1)||(c2=='>'&&m==3))//逆时针
{
printf("ccw\n");
continue;
}
else//都有可能
{
printf("undefined\n");
}
}//重复四个判断即可；
else if(c1=='>')
{
if((c2=='v'&&m==1)||(c2=='^'&&m==3))
{
printf("cw\n");
continue;
}
else if((c2=='^'&&m==1)||(c2=='v'&&m==3))
{
printf("ccw\n");
continue;
}
else
{
printf("undefined\n");
}
}
else if(c1=='v')
{
if((c2=='<'&&m==1)||(c2=='>'&&m==3))
{
printf("cw\n");
continue;
}
else if((c2=='>'&&m==1)||(c2=='<'&&m==3))
{
printf("ccw\n");
continue;
}
else
{
printf("undefined\n");
}
}
else if(c1=='<')
{
if((c2=='^'&&m==1)||(c2=='v'&&m==3))
{
printf("cw\n");
continue;
}
else if((c2=='v'&&m==1)||(c2=='^'&&m==3))
{
printf("ccw\n");
continue;
}
else
{
printf("undefined\n");
}
}
}
return 0;
}