Codeforces Round #426 (Div. 2) The Useless Toy
2017-07-31 11:25
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题意:东南西北四个方向,每次转到上一个或者下一个位置。现在告诉你 转的次数 和 起点 终点 问你是经过顺时针得到的还是逆时针得到的。
思路:hash一下,然后算出起点顺时针和逆时针转n次的终点,判断一下关系就OVER了
思路:hash一下,然后算出起点顺时针和逆时针转n次的终点,判断一下关系就OVER了
#include<bits/stdc++.h> using namespace std; int judge(char ch) // hash { if(ch=='^') return 1; if(ch=='>') return 2; if(ch=='v') return 3; if(ch=='<') return 0; } int main(void) { int n; char s1[2]; char s2[2]; scanf("%s%s",s1,s2); //printf("%s %s",s1,s2); cin >> n; int a=judge(s1[0]); //起点hash int b=judge(s2[0]); // 终点hash int x=(a+n)%4; //顺时针得到的终点 int y=(((a-n)%4)+4)%4;//逆时针得到的终点 //printf("x=%d y=%d\n",x,y); if(x==y) { printf("undefined\n"); return 0; } else if(x==b) { printf("cw\n"); return 0; } else if(y==b) { printf("ccw\n"); return 0; } }
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