Simple implementation of wuziqi in C Language

Wuziqi implementationCode

# Include <stdio. h> # define M 9 # define N 9 void print (int A [m] [N]) {int I, j; for (I = 0; I <m; I ++) {for (j = 0; j <n; j ++) printf ("% d", a [I] [J]); printf ("\ n") ;}} int check (int A [m] [N]) {int I, j; for (I = 0; I <M-5; I ++) {for (j = 0; j <N-5; j ++) {If (A [I] [J] = 1) & (A [I + 1] [J] = 1) & (A [I + 2] [J] = 1) & (A [I + 3] [J] = 1) & (A [I + 4] [J] = 1 )) {printf ("player1 win \ n"); Return 0; break;} else if (A [I] [J] = 1) & (A [I] [J + 1] = 1) & (A [I] [J + 2] = 1) & (A [I] [J + 3] = 1) & (A [I] [J + 4] = 1 )) {printf ("player1 win \ n"); Return 0; break;} else if (A [I] [J] = 1) & (A [I + 1] [J + 1] = 1) & (A [I + 2] [J + 2] = 1) & (A [I + 3] [J + 3] = 1) & (A [I + 4] [J + 4] = 1 )) {printf ("player1 win \ n"); Return 0; break;} else if (A [M-I] [J] = 1) & (A [M-i-1] [J + 1] = 1) & (A [M-i-2] [J + 2] = 1) & (A [M-i-3] [J + 3] = 1) & (A [M-i-4] [J + 4] = 1 )) {printf ("player1 win \ n"); Return 0; break;} If (A [I] [J] = 2) & (A [I + 1] [J] = 2) & (A [I + 2] [J] = 2) & (A [I + 3] [J] = 2) & (A [I + 4] [J] = 2 )) {printf ("player2 win \ n"); Return 0; break;} else if (A [I] [J] = 2) & (A [I] [J + 1] = 2) & (A [I] [J + 2] = 2) & (A [I] [J + 3] = 2) & (A [I] [J + 4] = 2 )) {printf ("player2 win \ n"); Return 0; break;} else if (A [I] [J] = 2) & (A [I + 1] [J + 1] = 2) & (A [I + 2] [J + 2] = 2) & (A [I + 3] [J + 3] = 2) & (A [I + 4] [J + 4] = 2 )) {printf ("player2 win \ n"); Return 0; break;} else if (A [M-I] [J] = 2) & (A [M-i-1] [J + 1] = 2) & (A [M-i-2] [J + 2] = 2) & (A [M-i-3] [J + 3] = 2) & (A [M-i-4] [J + 4] = 2 )) {printf ("player2 win \ n"); Return 0; break ;}}return 1 ;}int main () {int A [m] [N] = {0}, B [m] [N] = {0}; print (a); int I, J; int COUNT = 1, num = 0; while (count) {num ++; If (Num % 2 = 1) {printf ("player1 input :"); scanf ("% d", & I); scanf ("% d", & J); A [I-1] [J-1] = 1; print (); count = check (a); for (I = 0; I <m; I ++) for (j = 0; j <n; j ++) B [I] [J] = A [I] [J];} else if (Num % 2 = 0) {printf ("player2 input :"); scanf ("% d", & I); scanf ("% d", & J); B [I-1] [J-1] = 2; print (B ); count = check (B); for (I = 0; I <m; I ++) for (j = 0; j <n; j ++) A [I] [J] = B [I] [J] ;}} return 0 ;}

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