SDNU——8题——G - The Useless Toy

最新推荐文章于 2022-11-15 18:02:28 发布

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最新推荐文章于 2022-11-15 18:02:28 发布

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本文介绍了一种通过记录旋转玩具Caramel Spinner起始位置和结束位置的方法来判断其旋转方向的算法实现。该玩具以V形焦糖的形式存在，可进行顺时针或逆时针旋转，并在每个方向停留一秒。文章提供了一个使用C++编写的解决方案，利用map存储四个位置字符及其对应的数值，通过计算确定旋转方向。

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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.

Example
Input
^ >
1
Output
cw
Input
< ^
3
Output
ccw
Input
^ v
6
Output

undefined

自从那次更新到F题后就一直没来的及补全，现在终于缓过气来了，虽然还没有完全消化学过的内容，但是终于有时间整理整理这段时间做的题啦

这一道题我还有印象

觉得还是比较简单的

如果能够想到用map那接下来就是分轻松了

接下来就是看看顺时针行不行，逆时针行不行，如果都行的情况，注意数不能超过4（%4）

#include<cstdio>
#include<iostream>
#include<string>
#include<map>
using namespace std;
int main()
{
	char s1,s2;
	map<char,int> mp;
	int n;
	while(cin>>s1>>s2)
	{
		int n;
		cin>>n;
		n %= 4;
		mp['^'] = 0;
		mp['>'] = 1;
		mp['v'] = 2;
	    mp['<'] = 3;
	    if((mp[s1] + n) % 4 == mp[s2] && (mp[s1] - n + 4) % 4 == mp[s2])
	    {
	    	cout<<"undefined"<<endl;
		}
		else if((mp[s1] + n) % 4 == mp[s2])
		{
			cout<<"cw"<<endl;
		}
		else if((mp[s1] - n + 4) % 4 == mp[s2])
		{
			cout<<"ccw"<<endl;
		}
	}


	return 0;
}

endl；