Hdu-4452-Running Rabbits

A square field of N * N is divided into 1*1 grids. There are two rabbits. Tom is in the upper left corner (1, 1) and Jerry is in the lower right corner (N, N ), they can go in four directions in the Southeast and northwest regions, but they cannot go out of order. Each has its own speed. When they meet each other, the two rabbits exchange the direction of movement and reverse the way they touch the border, they also have their own turn to the left cycle, the cycle is one to turn to the left (but at this time the two rabbits do not execute this cycle to the left), ask the coordinates of the two rabbits after K hours. --> Just follow the instructions. Boundary processing: If N = 5, right or bottom bounds: x (or y) directly modulo 2 * N-21234543212345432123 ...... Cycle is 2*5-2, save to array a; left or upper bounds: x (or y) to 1-x (OR 1-y) and then modulo 2 * N-2 ...... The 123454321234543212345 period is 2*5-2 and is saved to array B. I didn't notice speed (1 ≤ s <N). I thought it would change the direction multiple times and write more errors ...... In fact, it's quite difficult ...... [Cpp] # include <cstdio> # include <algorithm> using namespace std; int N, a [50], B [50]; void turn (char & c) // turn left {switch (c) {case 'E': c = 'n'; break; case 's': c = 'E'; break; case 'W': c = 's'; break; case 'N': c = 'W'; break;} void moveto (char & c, int S, int & x, int & y) // step {switch (c) {case 'E': {y + = s; if (y> N) {y = a [y % (2 * N-2)]; c = 'W';} break;} case 's': {x + = S; if (x> N) {x = A [x % (2 * N-2)]; c = 'n';} break;} case 'W': {y-= s; if (y <1) {y = B [(1-y) % (2 * N-2)]; c = 'E';} break;} case 'N': {x-= s; if (x <1) {x = B [() % (2 * N-2)]; c = 's';} break ;}} int main () {int K, T_s, T_t, J_s, J_t, I; char T_c, J_c; while (~ Scanf ("% d", & N) {if (! N) return 0; scanf ("\ n % c % d", & T_c, & T_s, & T_t ); scanf ("\ n % c % d", & J_c, & J_s, & J_t); scanf ("% d", & K); int T_x = 1, t_y = 1, J_x = N, J_y = N; a [0] = 2; for (I = 1; I <= N; I ++) a [I] = I; for (I = N + 1; I <= 2 * N-3; I ++) a [I] = 2 * N-I; B [0] = 1; for (I = 1; I <= N-1; I ++) B [I] = I + 1; for (I = N; I <= 2 * N-3; I ++) B [I] = 2 * N-i-1; for (I = 1; I <= K; I ++) {moveto (T_c, t_s, T_x, T_y); moveto (J_c, J_s, J_x, J_y); if (T_x = J_x & T_y = J_y) swap (T_c, J_c ); else {if (I % T_t = 0) turn (T_c); if (I % J_t = 0) turn (J_c );}} printf ("% d \ n", T_x, T_y); printf ("% d \ n", J_x, J_y);} return 0 ;}