// Created by Siyabonga Brian // Created by Siyabonga Brian /* Position and direction on grid Movement is in the following rules: 1. At a white square on my grid (indicated by a zero), turn 90 degrees right, move forward one square. 2. At a black square on my grid (indicated by a one), turn 90 degrees left, move forward one square. Basically I need to get the new location on my grid after I was on a certain location. Code needs to accept the colour of the cell I'm currently at, as well as position and direction I am facing and computes the cell's new colour, new position and direction after taking a single step. INPUT: The first line of input is a single integer, representing the colour of the cell. The next line consists of two integers m and n and a character d, separated by a single space, specifying the row and column location of the ant, and the direction it is facing (either N, S, E, or W for the four cardinal directions). for example. Input: 1 3 2 E Output: 0 2 2 N https://code.sololearn.com/cyW9P3xVI3Bd */ #include #include using namespace std; int main() { // vector>grid = { "00000", "10010", "11100", "00000", "00000" }; int r = 5, c = 5; char grid[r][c+1] = { "00000", "10010", "11100", "00000", "00000" }; /* cin >> r >> c; // get row and column for(int i = 0; i < r; i++){ vectortemp; char input; for(int j = 0; j < c; j++){ cin >> input; temp.push_back(input); } grid.push_back(temp); } for(int i = 0; i < r; i++){ for(int j = 0; j < c; j++){ cout << grid[i][j]; } cout << endl; } */ char D[] = "NWSE"; int color, m, n; // color, row, column char d; // orientation cin >> color >> m >> n >> d; // identify index of d in D: int di = -1; while(toupper(d) != D[++di]); // calculate output: di = (di+3 + color*2) % 4; m += (di-1)%2; n += (di-2)%2; d = D[di]; //printf("%c\n%d %d %c", grid[m][n], m, n, d); // for 0-based input coordinates printf("%c\n%d %d %c", grid[m-1][n-1], m, n, d); // for 1-based input coordinates return 0; }

grid code 4