n=4
a=[[i*n+j+1 for j in range(n)] for i in range(n)]
for i in range(n//2):
for j in range(1, n, 2):
a[i][j], a[n-i- 1][n-j- 1]=a[n-i- 1][n-j- 1], a[i][j]
则程序执行后, a[1][1]和 a[2][0]的值分别为( )
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