我被要求编写一个内存游戏,有一些细节,其中第一次显示 Revel 的字母,然后,如果用户在第二个提示中猜测相应的匹配点,那么棋盘应该保持这样,直到用户完成游戏(通过猜测所有正确的匹配点),这是一个 2x2 网格的示例
Your program:
* *
* *
Enter a pair of integers in the range [1, 2]
Player: 1 1
Your program:
A *
* *
(then it clears the screen and displays)
* *
* *
Enter another pair of integers in the range [1, 2]
Player: 2 1
Your program:
* *
C *
(then it clears screen and displays)
* *
* *
Enter a pair of integers in the range [1, 2]
Player: 1 2
Your program:
* C
* *
(then it clears screen and displays)
* *
* *
Enter another pair of integers in the range [1, 2]
Player: 2 1
Your program:
* *
C *
(then it clears the screen and displays)
* C
C *
Enter a pair of integers in the range [1, 2]
Player: 1 1
Your program:
A C
C *
(then it clears the screen and displays)
* C
C *
Enter another pair of integers in the range [1, 2]
Player: 1 1
Your program:
A C
C *
(then it clears the screen and displays)
* C
C *
Enter a pair of integers in the range [1, 2]
Player: 1 1
Your program:
A C
C *3
(then it clears the screen and displays)
* C
C *
Enter another pair of integers in the range [1, 2]
Player: 2 2
Your program:
A C
C A
CONGRATULATIONS. YOU SUCCEEDED
我需要一个 4x4,我知道如何显示正确的匹配,但我似乎无法存储新板,所以用户看到最新的板,我无法理解它......
char board[4][4] = { {'A','B','A','D'},{'C','E','H','G'},{'B','D','G','C'},{'F','H','F','E'} };
int i, j, row, column, row2, column2;
char boardmatch[4][4];
int tempX,tempY;
for(tempX = 0; tempX < 4; tempX++){
for(tempY = 0; tempY < 4; tempY++){
boardmatch[tempX][tempY] = 0;
}
}
for (i=0; i < 4 ; i++){
for (j=0; j<4; j++){
printf("* ");
}
printf("\n");
}
do {
printf("\nEnter a pair of integers in the range [1, 4]: ");
scanf("%d %d", &row, &column);
row--;
column--;
printf("\n");
for (i=0; i < 4 ; i++){
for (j=0; j < 4 ; j++){
if ( i == row && j == column){
printf("%c ", board[row][column]);
}else{
printf("* ");
}
}
printf("\n");
}
printf("\n");
system("pause");
system("cls");
for (i=0; i < 4 ; i++){
for (j=0; j<4; j++){
printf("* ");
}
printf("\n");
}
printf("\nEnter another pair of integers in the range [1, 4]: ");
scanf("%d %d", &row2, &column2);
row2--;
column2--;
printf("\n");
for (i=0; i < 4 ; i++){
for (j=0; j < 4 ; j++){
if (i == row2 && j == column2){
printf("%c ", board[row2][column2]);
}else{
printf("* ");
}
}
printf("\n");
}
system("pause");
system("cls");
if(board[row][column]==board[row2][column2]){
boardmatch[row][column] = 1;
boardmatch[row2][column2] = 1;
}
for (i=0; i < 4 ; i++){
for (j=0; j<4; j++){
if (boardmatch[i][j] == 1){
printf("%c ", board[row2][column2]);
}else{
printf("* ");
}
}
printf("\n");
}
printf("\n");
system("pause");
system("cls");
}while(1);
system("PAUSE");
return 0;
}
最佳答案
您需要另一个用于板的阵列。
它只为每个单元格保存一个位,意思是“找到”或“转变”。
使用它(和原始板)来显示板。
仅显示已找到/打开的单元格。
对于玩家回合,当某些单元格可能被转回时,只需记住哪个单元格被转回,以便您可以将其转回未发现/未转回。该数组开始时全部未翻转,当全部翻转时游戏结束。
(您也可以使用结构将所有内容放入一个数组中。)
关于c - 如何存储掩码数组,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/17871730/