Write up

1.

题目是

p = 1325465431
q = 152317153
e = 65537
计算出d,将d用MD5加密后包裹NSSCTF{}提交

这是一道rsa和MD5加密题目

import hashlib
p = 1325465431
q = 152317153
e = 65537
n = p * q
print(f"n={n}")
phi_n = (p - 1) * (q - 1)
d = pow(e, -1, phi_n)#e对phi_n的模逆
print(f"d={d}")
d_str = str(d)
md5_hash = hashlib.md5(d_str.encode()).hexdigest()
result = f"NSSCTF{{{md5_hash}}}"
print(result)

n=201891120849837943
d=43476042047970113
NSSCTF{08bb8fb628da85923e5734a75ac19ffe}

所以答案就是NSSCTF{08bb8fb628da85923e5734a75ac19ffe}

2.

题目是：

dict:{0: 'J', 1: 'K', 2: 'L', 3: 'M', 4: 'N', 5: 'O', 6: 'x', 7: 'y', 8: 'U', 9: 'V', 10: 'z', 11: 'A', 12: 'B', 13: 'C', 14: 'D', 15: 'E', 16: 'F', 17: 'G', 18: 'H', 19: '7', 20: '8', 21: '9', 22: 'P', 23: 'Q', 24: 'I', 25: 'a', 26: 'b', 27: 'c', 28: 'd', 29: 'e', 30: 'f', 31: 'g', 32: 'h', 33: 'i', 34: 'j', 35: 'k', 36: 'l', 37: 'm', 38: 'W', 39: 'X', 40: 'Y', 41: 'Z', 42: '0', 43: '1', 44: '2', 45: '3', 46: '4', 47: '5', 48: '6', 49: 'R', 50: 'S', 51: 'T', 52: 'n', 53: 'o', 54: 'p', 55: 'q', 56: 'r', 57: 's', 58: 't', 59: 'u', 60: 'v', 61: 'w', 62: '+', 63: '/', 64: '='}

chipertext:
FlZNfnF6Qol6e9w17WwQQoGYBQCgIkGTa9w3IQKw

这是新建了一个base字典需要我们自己更换字典进行解密

这个字典一个一个地提取字母太麻烦了，我们直接用代码提取一下

original_list = ['J', 'K', 'L', 'M', 'N', 'O', 'x', 'y', 'U', 'V', 'z', 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', '7', '8', '9', 'P', 'Q', 'I', 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'W', 'X', 'Y', 'Z', '0', '1', '2', '3', '4', '5', '6', 'R', 'S', 'T', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', '+', '/', '=']
new_string = "".join(original_list)
print(new_string)

这是提取出来的新字典：

JKLMNOxyUVzABCDEFGH789PQIabcdefghijklmWXYZ0123456RSTnopqrstuvw+/=

#可以更改的字典
s = "JKLMNOxyUVzABCDEFGH789PQIabcdefghijklmWXYZ0123456RSTnopqrstuvw+/="

def My_base64_encode(inputs):
    # 将字符串转化为2进制
    bin_str = []
    for i in inputs:
        x = str(bin(ord(i))).replace('0b', '')
        bin_str.append('{:0>8}'.format(x))
    # print(bin_str)
    # 输出的字符串
    outputs = ""
    # 不够三倍数，需补齐的次数
    nums = 0
    while bin_str:
        # 每次取三个字符的二进制
        temp_list = bin_str[:3]
        if (len(temp_list) != 3):
            nums = 3 - len(temp_list)
            while len(temp_list) < 3:
                temp_list += ['0' * 8]
        temp_str = "".join(temp_list)
        # print(temp_str)
        # 将三个8字节的二进制转换为4个十进制
        temp_str_list = []
        for i in range(0, 4):
            temp_str_list.append(int(temp_str[i * 6:(i + 1) * 6], 2))
        # print(temp_str_list)
        if nums:
            temp_str_list = temp_str_list[0:4 - nums]

        for i in temp_str_list:
            outputs += s[i]
        bin_str = bin_str[3:]
    outputs += nums * '='
    print("Encrypted String:\n%s " % outputs)

def My_base64_decode(inputs):
    # 将字符串转化为2进制
    bin_str = []
    for i in inputs:
        if i != '=':
            x = str(bin(s.index(i))).replace('0b', '')
            bin_str.append('{:0>6}'.format(x))
    # print(bin_str)
    # 输出的字符串
    outputs = ""
    nums = inputs.count('=')
    while bin_str:
        temp_list = bin_str[:4]
        temp_str = "".join(temp_list)
        # print(temp_str)
        # 补足8位字节
        if (len(temp_str) % 8 != 0):
            temp_str = temp_str[0:-1 * nums * 2]
        # 将四个6字节的二进制转换为三个字符
        for i in range(0, int(len(temp_str) / 8)):
            outputs += chr(int(temp_str[i * 8:(i + 1) * 8], 2))
        bin_str = bin_str[4:]
    print("Decrypted String:\n%s " % outputs)

num = input("输入1进行加密，输入2进行解密:\n")
if (num == "1"):
    input_str = input("输入需要加密的字符串: \n")
    My_base64_encode(input_str)
else:
    input_str = input("输入需要解密的字符串: \n")
    My_base64_decode(input_str)

最后用NSSCTF包裹得到NSSCTF{D0_Y0u_kNoW_Th1s_b4se_map}