题目来源与2023河南省ccpc
ls = ['''
........
........
.0000000
.0.....0
.0.....0
.0.....0
.0.....0
.0.....0
.0000000
........
''', '''........
........
.......1
.......1
.......1
.......1
.......1
.......1
.......1
........''', '''.........
.........
.2222222.
.......2.
.......2.
.2222222.
.2.......
.2.......
.2222222.
.........''', '''........
........
.3333333
.......3
.......3
.3333333
.......3
.......3
.3333333
........''', '''........
........
.4.....4
.4.....4
.4.....4
.4444444
.......4
.......4
.......4
........''', '''........
........
.5555555
.5......
.5......
.5555555
.......5
.......5
.5555555
........''', '''........
........
.6666666
.6......
.6......
.6666666
.6.....6
.6.....6
.6666666
........''', '''........
........
.7777777
.......7
.......7
.......7
.......7
.......7
.......7
........''', '''........
........
.8888888
.8.....8
.8.....8
.8888888
.8.....8
.8.....8
.8888888
........''', '''........
........
.9999999
.9.....9
.9.....9
.9999999
.......9
.......9
.9999999
........
''']
s = ls[1].split('\n')
s1 = ls[2].split('\n')
ss = '\n'.join(''.join(line) for line in zip(s, s1))
print(ss)
