题意：给你一个HxW的矩阵，每个点是一个指令，根据指令进行一系列操作。
题解：模拟
#include#include using namespace std;const int maxn = 101;char G[maxn][maxn];int dx[] = {-1,0,1, 0};int dy[] = { 0,1,0,-1};struct pointer{ int r,c; int dir; void Move(){ r += dx[dir]; c += dy[dir]; } char read(){ return G[r][c]; }}p; int h,w;const int up_bound = 1e5;const int TimeLim = 1e6;const int MaxN = 1e5+5;int readLim;int readList[MaxN];void execute(){ p.r = p.c = 0; int Register[26] = { 0}; int cur = 0; char op = p.read(); p.dir = 1; int step = 1; int ReadTimes = 0; while(op!='#'){ if('A'<=op&&op<='Z'){ swap(Register[op-'A'],cur); }else switch(op){ case '^':{ p.dir = 0; break; } case '>':{ p.dir = 1; break; } case 'v':{ p.dir = 2; break; } case '<':{ p.dir = 3; break; } case '?':{ if(ReadTimes>=readLim){ cur = readList[readLim-1]; }else { cur = readList[ReadTimes]; ReadTimes++; } break; } case '!':{ printf("%d\n",cur); cur = 0; break; } case '+':{ cur++; if(abs(cur)>up_bound){ printf("OVERFLOW ERROR\n"); return; } break; } case '-':{ cur--; if(abs(cur)>up_bound){ printf("OVERFLOW ERROR\n"); return; } break; } case '@':{ if(cur){ p.dir = (p.dir+1)%4; } else { p.dir = (p.dir+3)%4; } break; } } step++; if(step>TimeLim) { printf("TIME LIMIT EXCEEDED\n"); return; } p.Move(); if(p.r<0||p.r>=h||p.c<0||p.c>=w) { printf("RUNTIME ERROR\n"); return; } op = p.read(); }}int main(){ scanf("%d%d",&h,&w); for(int i = 0; i < h; i++) scanf("%s",G[i]); scanf("%d",&readLim); for(int i = 0; i < readLim; i++){ scanf("%d",readList+i); } execute(); return 0;}