Codeforces

A. The Useless Toy

题目链接

分类：模拟、map应用

1.题意概述

• 给你两个字符和旋转次数，问从第一个字符到第二个字符是通过顺时针还是逆时针旋转还是两种方向都行？

2.解题思路

• 考虑map映射4种状态，显然可以根据旋转的次数来确定它是否都能到达即可，具体细节参见代码。

3.AC代码

#include <bits/stdc++.h>using namespace std;#define lson root << 1#define rson root << 1 | 1#define lent (t[root].r - t[root].l + 1)#define lenl (t[lson].r - t[lson].l + 1)#define lenr (t[rson].r - t[rson].l + 1)#define eps 1e-6#define e exp(1.0)#define pi acos(-1.0)#define fi first#define se second#define pb push_back#define mp make_pair#define SZ(x) ((int)(x).size())#define All(x) (x).begin(),(x).end()#define rep(i,a,n) for (int i=a;i<n;i++)#define per(i,a,n) for (int i=n-1;i>=a;i--)#define Close() ios::sync_with_stdio(0),cin.tie(0)#define INF 0x3f3f3f3f#define maxn 1000010#define N 1111typedef vector<int> VI;typedef pair<int, int> PII;typedef long long ll;typedef unsigned long long ull;const int mod = 1e9 + 7;/* head */int main() {#ifndef ONLINE_JUDGE    freopen("in.txt", "r", stdin);    freopen("out.txt", "w", stdout);    long _begin_time = clock();#endif    char a, b;    scanf("%c %c", &a, &b);    map<char, int> mp;    mp['^'] = 4;    mp['v'] = 2;    mp['<'] = 1;    mp['>'] = 3;    int n;    scanf("%d", &n);    n %= 4;    if((n & 1) == 0)         puts("undefined");    else if((mp[a] - mp[b] + 4) % 4 == n)        puts("cw");    else puts("ccw");#ifndef ONLINE_JUDGE    long _end_time = clock();    printf("time: %ld ms\n", _end_time - _begin_time);#endif    return 0;}