题解 | #矩阵交换#

int main()

{

int n = 0, m = 0;

scanf("%d%d", &n, &m);

int arr[10][10] = { 0 };

int i = 0, j = 0;

for (i = 0; i < n; i++)

{

for (j = 0; j < m; j++)

{

scanf("%d", &arr[i][j]);

}

}

int k = 0;

scanf("%d", &k);

int a = 0, b = 0;

char t=' ';

//循环k次

while(k)

{

getchar();//将缓冲区内空格或回车清除

scanf("%c%d%d",&t,&a, &b);

//行转换

//每次换行或列，需要相对应的减一

if (t == 'r')

{

for (j = 0; j < m; j++)

{

int temp = arr[a-1][j];

arr[a-1][j] = arr[b-1][j];

arr[b-1][j] = temp;

}

}

//列转换

else if (t == 'c')

{

for (i = 0; i < n; i++)

{

int temp = arr[i][a-1];

arr[i][a-1] = arr[i][b-1];

arr[i][b-1] = temp;

}

}

//其它字符直接跳过

else

{

break;

}

k--;

}

//输出

for (i = 0; i < n; i++)

{

for (j = 0; j < m; j++)

{

printf("%d ", arr[i][j]);

}

printf("\n");

}

return 0;

}