题目如上描述，思路是如果小人的朝向与查询的方向相同就减，不同就加，然后直接模拟。 毕竟是day1T1还比较简单。 附AC代码：
#include<cstdio> #include<iostream> #include<ctime> #include<cstdlib> #include<algorithm> #include<cmath> #include<cstring> #include<string> using namespace std; const int maxn = 1e5+5; int n,m; int dic,len; int tws[maxn]; char mz[maxn][15]; template <class T> inline void read(T &xx) { xx = 0; T flag = 1; char ch = (char)getchar(); while(ch<'0' || ch>'9') { if(ch == '-') flag = -1; ch = (char)getchar(); } while(ch>='0' && ch<='9') { xx = (xx<<1) + (xx<<3) + ch - '0'; ch = (char)getchar(); } xx *= flag; } int main() { freopen("toy.in","r",stdin); freopen("toy.out","w",stdout); int now = 1,len; read(n); read(m); for(int i = 1; i <= n; i++) { read(tws[i]); scanf("%s",mz[i]); } while(m--) { read(dic); read(len); if(dic == tws[now]) { if(now > len) now -= len; else now = n-len+now; } else if(dic != tws[now]) { if(now+len > n) now = len+now-n; else now = len+now; } } for(int i = 0; i < strlen(mz[now]); i++) printf("%c",mz[now][i]); return 0; }