编写题解 2900: 螺旋加密

解题思路:

注意事项:

参考代码:

d = input().split()

if(len(d)>3):

    for i in range(3,len(d)):

        d[2] += ' '+d[i]

if(len(d)<3):

    d.append(' ')

r,c,ts = int(d[0]),int(d[1]),d[2]

m = [['0']*c for _ in range(r)]

#转化字符位5位二进制

def schr(tc):

    digit = 0

    if tc != ' ':

        digit = ord(tc) - 64

    t = ['0','0','0','0','0']

    k = []

    while digit > 0:

        k += str(digit%2)

        digit //= 2

    t[5-len(k):5] = k[::-1]

    return t

s = ''

for i in ts:

    s += ''.join(schr(i))

if len(s) < r*c:

    for i in range(len(s),r*c):

        s += '0'

up ,down,left,right = 0,r-1,0,c-1

si = 0

while up<=down and left<=right and si < len(s):

    #从左到右

    for i in range(left,right+1):

        if si < len(s):

            m[up][i] = s[si]

            si += 1

    up += 1

    #从上到下

    for i in range(up,down+1):

        if si < len(s):

            m[i][right] = s[si]

            si += 1

    right -= 1

    #从右到左

    for i in range(left,right+1)[::-1]:

        if si < len(s):

            m[down][i] = s[si]

            si += 1

    down -= 1

    #从下到上

    for i in range(up,down+1)[::-1]:

        if si < len(s):        

            m[i][left] = s[si]

            si += 1

    left += 1

for i in range(r):

    for j in range(c):

        print(m[i][j],end='')