UNCTF2019：BABYRSA

下载附件，解压后有一个加密文件和rsa.py代码如下：

flag = open('flag.txt','r').read()
N = 221
e = 5
enc = b''

for i in flag:
    enc += bytes([pow(ord(i),e,N)])

encrypt = open('encrypt','wb')
encrypt.write(enc)
encrypt.close()

　　知道加密脚本我们反向编写脚本获取flag：

flag=""
N = 221
e = 5
with open('encrypt','rb') as fp:
    encrypt=fp.read()
encrypt=bytes.decode(encrypt)
for i in encrypt:
    for x in range(0,255):
        if pow(x,e,N)==ord(i):
            flag+=chr(x)
            break
print(flag)

得到flag为：UNCTF{10023493ff87a9f246eb09d4f573e060}