回形取数--简单

解题思路:

注意事项:

参考代码:

#include<iostream>

using namespace std;

int main()

{

int m, n;

cin >> m >> n;

int num[500][500];

for (int i = 0; i < m; i++)

for (int j = 0; j < n; j++)

{

cin >> num[i][j];

}

int i = 0, j = 0;

int time = 0;

int sum = m * n - 1;

cout << num[0][0];

while (true)

{

for (i++; i < (m - time); i++)

{

cout << " " << num[i][j];

sum--;

}

i--;

if (sum == 0)

break;

for (j++; j < (n - time); j++)

{

cout << " " << num[i][j];

sum--;

}

j--;

if (sum == 0)

break;

for (i--; i >= time; i--)

{

cout << " " << num[i][j];

sum--;

}

i++;

if (sum == 0)

break;

for (j--; j >= time + 1; j--)

{

cout << " " << num[i][j];

sum--;

}

j++;

if (sum == 0)

break;

time++;

}

return 0;

}