swpu新生赛ctf wp

WEB:gift_F12

没啥好说的直接F12得了

NSSCTF{We1c0me_t0_WLLMCTF_Th1s_1s_th3_G1ft}

RE

简简单单的解密

import base64, urllib.parse

def e():

    key = "HereIsFlagggg"

    flag = "xxxxxxxxxxxxxxxxxxx"

    s_box = list(range(256))

    j = 0

    # 打乱s_box 和flag无关

    for i in range(256):

        j = (j + s_box[i] + ord(key[i % len(key)])) % 256

        s_box[i], s_box[j] = s_box[j], s_box[i]

    res = []

    i = j = 0

    for s in flag:

        i = i + 1

        j = (j + s_box[i]) % 256

        s_box[i], s_box[j] = s_box[j], s_box[i]

        t = (s_box[i] + s_box[j]) % 256

        k = s_box[t]

        res.append(chr(ord(s) ^ k))

    enc = "".join(res)

    #原来这里有个b64加密再解密 就等于没变 所以改了一下

    enc = urllib.parse.quote(enc)

    print(enc)

def d():

    enc = "%C2%A6n%C2%87Y%1Ag%3F%C2%A01.%C2%9C%C3%B7%C3%8A%02%C3%80%C2%92W%C3%8C%C3%BA"

    enc = urllib.parse.unquote(enc)#倒着往回做呗

    key = "HereIsFlagggg"

    s_box = list(range(256))

    j = 0

    r=[]

    # 获得s_box

    for i in range(256):

        j = (j + s_box[i] + ord(key[i % len(key)])) % 256

        s_box[i], s_box[j] = s_box[j], s_box[i]

    i,j=0,0

    for s in enc:

        i = i + 1

        j = (j + s_box[i]) % 256

        s_box[i], s_box[j] = s_box[j], s_box[i]

        t = (s_box[i] + s_box[j]) % 256

        k = s_box[t]

        r.append(chr(ord(s)^k))#基本直接复制就行 异或的逆运算就是再异或一次

    print("".join(r))

if __name__ == '__main__':

    e()

    d()

脚本运行得到结果

NSSCTF{REAL_EZ_RC4}

简简单单的逻辑

先上脚本

def e():

    flag = 'xxxxxxxxxxxxxxxxxx'

    list = [47, 138, 127, 57, 117, 188, 51, 143, 17, 84, 42, 135, 76, 105, 28, 169, 25]

    result = ''

    for i in range(len(list)):

        key = (list[i] >> 4) + ((list[i] & 0xf) << 4)

        a = hex(ord(flag[i]) ^ key)

        result += str(a)[2:].zfill(2)

        #分析得知 这是16进制补全两位 所以是一个flag字符对应两位

    print(result)

def d():

    list = [47, 138, 127, 57, 117, 188, 51, 143, 17, 84, 42, 135, 76, 105, 28, 169, 25]

    result = "bcfba4d0038d48bd4b00f82796d393dfec"

    nums = []

    #截取 转换 得到原来十进制

    for i in range(0,len(result),2):

        nums.append("0x" + result[i:i+ 2])

    for index in range(len(nums)):

        nums[index]=int(nums[index],16)

     #直接加密里面复制key的计算 然后再异或一次即可

    for i in range(len(list)):

        key = (list[i] >> 4) + ((list[i] & 0xf) << 4)

        print(chr(nums[i]^key),end="")

if __name__ == '__main__':

    # e()

    d()

这看脚本就行了

NSSCTF{EZEZ_RERE}

非常简单的逻辑题

def e():

    flag = 'xxxxxxxxxxxxxxxxxxxxx'

    s = 'wesyvbniazxchjko1973652048@$+-&*<>'

    result = ''

    for i in range(len(flag)):

        s1 = ord(flag[i]) // 17

        s2 = ord(flag[i]) % 17

        #就 没啥好讲的 一点点拆开打断点调试就能看懂计算过程

        #还是flag一个字符转换对应result的两个字符

        a = s[(s1 + i) % 34]

        b = s[-(s2 + i + 1) % 34]

        result += a + b

    print(result)

def d():

    result = 'v0b9n1nkajz@j0c4jjo3oi1h1i937b395i5y5e0e$i'

    s = 'wesyvbniazxchjko1973652048@$+-&*<>'

	#套这么多层循环 主要是我不知道 取余怎么更方便的算回去

    for x in range(-5, 5):

        for y in range(-5, 5):

            r = ""

            for j in range(0, len(result), 2):

                a = result[j]

                b = result[j + 1]

                #根据数学知识可得(

                s1 = int(s.index(a) + 34 * x - j / 2)

                s2 = -1*int(s.index(b) + 34 * y + j / 2 + 1)

                #去除没啥意义的结果

                if 32 < s1 * 17 + s2 < 130:

                    r += chr(s1 * 17 + s2)

                else:

                    break

            if len(r)==len('xxxxxxxxxxxxxxxxxxxxx'):

                print(r)

if __name__ == '__main__':

    # e()

    d()

NSSCTF{Fake_RERE_QAQ}

fakerondom

import random

def e():

    flag = 'xxxxxxxxxxxxxxxxxxxx'

    #别的都不重要 就是这个随机数种子

    #同样的随机数种子生成的一定是相同的数字

    random.seed(1)

    l = []

    for i in range(4):

        l.append(random.getrandbits(8))

    result = []

    for i in range(len(l)):

        random.seed(l[i])

        for n in range(5):

            result.append(ord(flag[i * 5 + n]) ^ random.getrandbits(8))

    print(result)

def d():

    result = [201, 8, 198, 68, 131, 152, 186, 136, 13, 130, 190, 112, 251, 93, 212, 1, 31, 214, 116, 244]

    random.seed(1)

    ran=[]

    t=[]

    for x in range(4):

        ran.append(random.getrandbits(8))

    l=ran[0:4]

    for i in range(len(l)):

        random.seed(l[i])

        for n in range(5):

        #异或题最好写了 基本原样复制代码就行

            print(chr(result[i * 5 + n]^ random.getrandbits(8)),end="")

if __name__ == '__main__':

    e()

    d()

NSSCTF{FakeE_random}

fakebase

def e():

    flag = 'xxxxxxxxxxxxxxxxxxx'

    s_box = 'qwertyuiopasdfghjkzxcvb123456#$'

    tmp = ''

    for i in flag:

        x = bin(ord(i))

        tmp += str(x)[2:].zfill(8)

    b1 = int(tmp, 2)

    s = ''

    while b1 // 31 != 0:

        c = b1 % 31

        s += s_box[c]

        b1 = b1 // 31

        print(b1)

    print(s)

def d():

    s = "u#k4ggia61egegzjuqz12jhfspfkay"

    # s="j3d4h1$6ggrouxktrgky5sxv6bk6i5"#测试

    s_box = 'qwertyuiopasdfghjkzxcvb123456#$'

    l = []

    for x in s:

        l.append(s_box.index(x))

    l=l[::-1]

    for n in range(0,31):

        for x in l:

            n=x+31*n

        t=str(bin(n))[2:]

        for j in range(0,len(t),8):

        	#这题最让我迷惑 我觉得我写的是对的

        	#但是拿上面的加密结果再解密出来的ascii值总差一倍

        	#但最后一个字符的ascii值又是对应的 所以只能除以2看看

        	#所以就 现在挺迷茫的

            print(chr(int(int(t[j:j+8],2)/2)),end="")

        print()

if __name__ == '__main__':

    # e()

    d()

NSSCTF{WHAt_BASe31}

astjs

经过查资料知道这是把js文件转换为ast 但是我不知道怎么转回去

解不出我太菜了

简单的逻辑

我是 san兵

我真看不出

出题人说少给条件了但可以琼剧

密码

带rsa的都不会萌新还没学呢

crypto6

var="************************************"

flag='NSSCTF{' + base64.b16encode(base64.b32encode(base64.b64encode(var.encode()))) + '}'

print(flag)

小明不小心泄露了源码，输出结果为：4A5A4C564B36434E4B5241544B5432454E4E32465552324E47424758534D44594C4657564336534D4B5241584F574C4B4B463245365643424F35485649534C584A5A56454B4D4B5049354E47593D3D3D，你能还原出var的正确结果吗？

会用base64库就行

import base64

x="4A5A4C564B36434E4B5241544B5432454E4E32465552324E47424758534D44594C4657564336534D4B5241584F574C4B4B463245365643424F35485649534C584A5A56454B4D4B5049354E47593D3D3D"

a=base64.b16decode(x)

a=base64.b32decode(a)

a=base64.b64decode(a)

print(a)

NSSCTF{5e110989-dc43-1bd3-00b4-9009206158fe}

Crypto7

69f7906323b4f7d1e4e972acf4abfbfc

直接md5解密

NSSCTF{md5yyds}

crypto8

介绍pyhton库:ciphey库! 人工智能 yyds

直接一把梭

终端执行命令

ciphey -t "73E-30U1&>V-H965S95]I<U]P;W=E<GT`"

这个题用了uuencode

crypto9

题干脚本直接python执行

letter_list = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'  # 字母表

# 根据输入的key生成key列表

def Get_KeyList(key):

    key_list = []

    for ch in key:

        key_list.append(ord(ch.upper()) - 65)

    return key_list

# 加密函数

def Encrypt(plaintext, key_list):

    ciphertext = ""

    i = 0

    for ch in plaintext:  # 遍历明文

        if 0 == i % len(key_list):

            i = 0

        if ch.isalpha():  # 明文是否为字母,如果是,则判断大小写,分别进行加密

            if ch.isupper():

                ciphertext += letter_list[(ord(ch) - 65 + key_list[i]) % 26]

                i += 1

            else:

                ciphertext += letter_list[(ord(ch) - 97 + key_list[i]) % 26].lower()

                i += 1

        else:  # 如果密文不为字母,直接添加到密文字符串里

            ciphertext += ch

    return ciphertext

# 解密函数

def Decrypt(ciphertext, key):

    plaintext = ""

    i = 0

    for ch in ciphertext:  # 遍历密文

        if 0 == i % len(key_list):

            i = 0

        if ch.isalpha():  # 密文为否为字母,如果是,则判断大小写,分别进行解密

            if ch.isupper():

                plaintext += letter_list[(ord(ch) - 65 - key_list[i]) % 26]

                i += 1

            else:

                plaintext += letter_list[(ord(ch) - 97 - key_list[i]) % 26].lower()

                i += 1

        else:  # 如果密文不为字母,直接添加到明文字符串里

            plaintext += ch

    return plaintext

if __name__ == '__main__':

    print("加密请按D,解密请按E:")

    user_input = input();

    while (user_input != 'D' and user_input != 'E'):  # 输入合法性判断

        print("输入有误!请重新输入:")

        user_input = input()

    print("请输入密钥:")

    key = input()

    while (False == key.isalpha()):  # 输入合法性判断

        print("输入有误!密钥为字母,请重新输入:")

        key = input()

    key_list = Get_KeyList(key)

    if user_input == 'D':

        # 加密

        print("请输入明文:")

        plaintext = input()

        ciphertext = Encrypt(plaintext, key_list)

        print("密文为:\n%s" % ciphertext)

    else:

        # 解密

        print("请输入密文:")

        ciphertext = input()

        plaintext = Decrypt(ciphertext, key_list)

        print("明文为:\n%s" % plaintext)

试了试密码NSS解密成功

Crypto10

当然你可以接着用这个

ciphey -t "AFFPGS{pbatenghyngvbaf!!!}"

NSSCTF{congratulations!!!}

ez_caesar

def caesar(plaintext):

    str_list = list(plaintext)

    i = 0

    while i < len(plaintext):

        if not str_list[i].isalpha():

            str_list[i] = str_list[i]

        else:

            a = "A" if str_list[i].isupper() else "a"

            str_list[i] = chr((ord(str_list[i]) - ord(a) + 5) % 26 + ord(a) or 5)

        i = i + 1

    return ''.join(str_list)

if __name__ == '__main__':

    small = [chr(i) for i in range(97, 123)]

    big = [chr(i) for i in range(65, 91)]

    alpha = small + big

    dic = {}

    for flag in alpha:

        str = caesar(flag)

        dic[str] = flag

    str = "SXXHYK{dtzmfajpstbhfjxfw}"

    for x in str:

        if x.isalpha():

            print(dic[x], end="")

        else:

            print(x, end="")

分析代码知道这是个静态加密所以直接先把所有字母对应的密文加到字典里面然后一一对应

NSSCTF{youhaveknowcaesar}

ez_rsa

为数不多我会写的rsa了

工具算个d 然后再md5加密即可

d=104550CB8E2144921

pigpig

看名字就知道了所以不想写

traditional

题目：

西方的二进制数学的发明者莱布尼茨，从中国的八卦图当中受到启发，演绎并推论出了数学矩

阵，

最后创造的二进制数学。二进制数学的诞生为计算机的发明奠定了理论基础。而计算机现在改

变

了我们整个世界，改变了我们生活，而他的源头却是来自于八卦图。现在，给你一组由八卦图

方位

组成的密文，你能破解出其中的含义吗？

震坤艮 震艮震 坤巽坤 坤巽震 震巽兑 震艮震 震离艮 震离艮

 格式：NSSCTF{}

我想说差不多得了(

八卦从内向外看 --代表0 —代表1

inp= "震坤艮 震艮震 坤巽坤 坤巽震 震巽兑 震艮震 震离艮 震离艮"

dic={"震":"100","离":"101","兑":"110","乾":"111","坤":"000","艮":"001","坎":"010","巽":"011"}

results= ""

for result in inp:

    if result in dic.keys():

        results+= dic[result][::-1]+""

    else:

        results+= " "

print(results)

results=results.split()

for index in range(0,len(results)):

    print(chr(int(results[index], 2)), end="")

写成二进制然后转十进制然后ascii码

NSSCTF{Da01sall}

misc

你喜欢osu吗？

我喜欢但不至强行安利真的(

OSU是一个windows的音游要是想玩这个谱的话可以去songs文件夹新建一个文件夹然后把两个文件扔进去你就能玩了

复制底下的铺面信息对就那一堆数字

224,224,0,5,0,0000

224,160,500,1,8,0000

224,160,1000,1,8,0000

224,160,1500,1,8,0000

224,160,2000,1,8,0000

224,224,2500,1,0,0000

224,160,3000,5,8,0000

224,160,3500,1,8,0000

224,224,4000,1,0,0000

224,224,4500,1,0,0000

224,160,5000,1,8,0000

224,224,5500,1,0,0000

224,224,6000,5,0,0000

224,224,6500,1,0,0000

224,224,7000,1,0,0000

224,160,7500,1,8,0000

224,224,8000,1,0,0000

224,160,8500,1,8,0000

224,160,9000,5,8,0000

224,160,9500,1,8,0000

224,224,10000,1,0,0000

224,160,10500,1,8,0000

224,224,11000,1,0,0000

224,160,11500,1,8,0000

224,224,12000,5,0,0000

224,160,12500,1,8,0000

224,160,13000,1,8,0000

224,160,13500,1,8,0000

224,224,14000,1,0,0000

224,224,14500,1,0,0000

224,160,15000,5,8,0000

224,160,15500,1,8,0000

224,224,16000,1,0,0000

224,160,16500,1,8,0000

224,160,17000,1,8,0000

224,224,17500,1,0,0000

224,160,18000,5,8,0000

224,160,18500,1,8,0000

224,160,19000,1,8,0000

224,160,19500,1,8,0000

224,224,20000,1,0,0000

224,160,20500,1,8,0000

224,224,21000,5,0,0000

224,160,21500,1,8,0000

224,160,22000,1,8,0000

224,160,22500,1,8,0000

224,160,23000,1,8,0000

224,160,23500,1,8,0000

224,224,24000,5,0,0000

224,160,24500,1,8,0000

224,160,25000,1,8,0000

224,224,25500,1,0,0000

224,224,26000,1,0,0000

224,160,26500,1,8,0000

224,224,27000,5,0,0000

224,160,27500,1,8,0000

224,224,28000,1,0,0000

224,160,28500,1,8,0000

224,160,29000,1,8,0000

224,224,29500,1,0,0000

224,160,30000,5,8,0000

224,224,30500,1,0,0000

224,160,31000,1,8,0000

224,160,31500,1,8,0000

224,224,32000,1,0,0000

224,160,32500,1,8,0000

224,160,33000,5,8,0000

224,224,33500,1,0,0000

224,160,34000,1,8,0000

224,224,34500,1,0,0000

224,224,35000,1,0,0000

224,160,35500,1,8,0000

224,224,36000,5,0,0000

224,160,36500,1,8,0000

224,160,37000,1,8,0000

224,224,37500,1,0,0000

224,160,38000,1,8,0000

224,160,38500,1,8,0000

224,224,39000,5,0,0000

224,224,39500,1,0,0000

224,224,40000,1,0,0000

224,160,40500,1,8,0000

224,224,41000,1,0,0000

224,160,41500,1,8,0000

224,160,42000,5,8,0000

224,160,42500,1,8,0000

224,160,43000,1,8,0000

224,160,43500,1,8,0000

224,224,44000,1,0,0000

224,160,44500,1,8,0000

224,160,45000,5,8,0000

224,160,45500,1,8,0000

224,224,46000,1,0,0000

224,160,46500,1,8,0000

224,224,47000,1,0,0000

224,160,47500,1,8,0000

224,224,48000,5,0,0000

224,160,48500,1,8,0000

224,160,49000,1,8,0000

224,224,49500,1,0,0000

224,160,50000,1,8,0000

224,160,50500,1,8,0000

224,160,51000,5,8,0000

224,160,51500,1,8,0000

224,224,52000,1,0,0000

224,160,52500,1,8,0000

224,160,53000,1,8,0000

224,160,53500,1,8,0000

224,160,54000,5,8,0000

224,224,54500,1,0,0000

224,224,55000,1,0,0000

224,160,55500,1,8,0000

224,224,56000,1,0,0000

224,160,56500,1,8,0000

224,224,57000,5,0,0000

224,160,57500,1,8,0000

224,160,58000,1,8,0000

224,160,58500,1,8,0000

224,160,59000,1,8,0000

224,160,59500,1,8,0000

224,224,60000,5,0,0000

224,160,60500,1,8,0000

224,160,61000,1,8,0000

224,160,61500,1,8,0000

224,224,62000,1,0,0000

224,160,62500,1,8,0000

224,160,63000,5,8,0000

224,160,63500,1,8,0000

224,224,64000,1,0,0000

224,160,64500,1,8,0000

224,160,65000,1,8,0000

224,224,65500,1,0,0000

224,160,66000,5,8,0000

224,160,66500,1,8,0000

224,160,67000,1,8,0000

224,160,67500,1,8,0000

224,224,68000,1,0,0000

224,160,68500,1,8,0000

224,160,69000,5,8,0000

224,224,69500,1,0,0000

224,160,70000,1,8,0000

224,160,70500,1,8,0000

224,160,71000,1,8,0000

224,224,71500,1,0,0000

224,224,72000,5,0,0000

224,160,72500,1,8,0000

224,160,73000,1,8,0000

224,224,73500,1,0,0000

224,160,74000,1,8,0000

224,224,74500,1,0,0000

224,160,75000,5,8,0000

224,160,75500,1,8,0000

224,224,76000,1,0,0000

224,160,76500,1,8,0000

224,224,77000,1,0,0000

224,160,77500,1,8,0000

224,160,78000,5,8,0000

224,160,78500,1,8,0000

224,160,79000,1,8,0000

224,160,79500,1,8,0000

224,224,80000,1,0,0000

224,160,80500,1,8,0000

224,160,81000,5,8,0000

224,224,81500,1,0,0000

224,160,82000,1,8,0000

224,224,82500,1,0,0000

224,224,83000,1,0,0000

224,160,83500,1,8,0000

224,224,84000,5,0,0000

224,160,84500,1,8,0000

224,160,85000,1,8,0000

224,160,85500,1,8,0000

224,160,86000,1,8,0000

224,160,86500,1,8,0000

224,224,87000,5,0,0000

224,160,87500,1,8,0000

224,224,88000,1,0,0000

224,160,88500,1,8,0000

224,224,89000,1,0,0000

224,224,89500,1,0,0000

224,224,90000,5,0,0000

224,160,90500,1,8,0000

224,160,91000,1,8,0000

224,224,91500,1,0,0000

224,224,92000,1,0,0000

224,160,92500,1,8,0000

224,224,93000,5,0,0000

224,160,93500,1,8,0000

224,224,94000,1,0,0000

224,160,94500,1,8,0000

224,224,95000,1,0,0000

224,224,95500,1,0,0000

224,224,96000,5,0,0000

224,160,96500,1,8,0000

224,224,97000,1,0,0000

224,224,97500,1,0,0000

224,224,98000,1,0,0000

224,224,98500,1,0,0000

224,160,99000,5,8,0000

224,160,99500,1,8,0000

224,224,100000,1,0,0000

224,160,100500,1,8,0000

224,224,101000,1,0,0000

224,160,101500,1,8,0000

然后观察嘛第二个数字 224转为0 160转为1 然后8位转换一个十进制当ascii码用最后来个翻转

import re

if __name__ == '__main__':

    with open("a.txt", "r")as file:

        s = file.read()

    s = s.split("\n")

    for index in range(0, len(s)):

        s[index] = s[index].split(",")[1]

        if s[index] == "224":

            s[index] = "0"

        else:

            s[index] = "1"

    s="".join(s)

    zero_one=re.findall(".{8}",s)

    r=[]

    for x in zero_one:

        r.append(chr(int(x,2)))

    print("".join(r[::-1]))

here_is_a_bug

真的我一定是非预期解

右键文件夹使用杀毒软件杀毒(

打开得知flag

NSSCTF{oh_you_catch_the_bug}

原来你也玩原神

异世相遇尽享美味!

file命令知道是个mp3

看了频谱文件不是摩斯啥的

用MP3stego解密密码留空就行

解密出一个txt文件看内容有PK

后缀名改为.zip

里面一个txt

NSSCTF{So_you_also_play_Genshin_impact}

Mooooooooooorse

AU打开你拿别的音频软件打开也行敲个摩斯码

解密就行了

文件已经忘了不想再敲了

Bill

看提示应该是个excel表格

后缀改为.xlsx 但是有密码

说不要被表象迷惑猜测可能是有隐写或者伪加密

foremost分离得到压缩包

后缀名再改xlsx打开文件

然后修改一下excel表格把所有的都换成中文大写数字之后把脚本拿出来

(上次DAS也有算钱的题...那个表格还比这个长

import openpyxl

import re

def trad_to_int(money):

    # 转换字典

    trad_dict = {

        '零':0,

        '壹':1,

        '贰':2,

        '叁':3,

        '肆':4,

        '伍':5,

        '陆':6,

        '柒':7,

        '捌':8,

        '玖':9,

        '拾':10,

        '佰':100,

        '仟':1000,

        '万':10000,

        '亿':100000000,

        '角':0.1,

        '分':0.01

    }

    trad = re.search(r"[零壹贰叁肆伍陆柒捌玖拾佰仟亿角分]+", money)

    if trad is not None:

        num = 0

        add = 0

        sum = 0

        for i in money:

            if i in ['零','壹','贰','叁','肆','伍','陆','柒','捌','玖']:

                add = trad_dict[i]

                sum = sum + add

            elif i in ['拾','佰','仟','亿','角','分']:

                num = add * trad_dict[i]

                sum = sum - add

                sum = sum + num

                add = num

            elif i == '万' or i == '亿':

                sum = sum * trad_dict[i]

        return float(sum)

    else:

        return money

def main():

    submoneyCN = ["", "拾", "佰", "仟"]

    numsCN = {"零": 0, "壹": 1, "贰": 2, "叁": 3, "肆": 4, "伍": 5, "陆": 6, "柒": 7, "捌": 8, "玖": 9}

    w = openpyxl.open("00000025.xlsx")

    ws = w.active

    # 获取第一列 即单件商品金额

    t = ws['A']

    t1=[]

    for x in range(1,len(t)):

        t1.append(trad_to_int(t[x].value))

    print(t1)

    # 获取第二列并处理

    t = ws['B']

    t2=[]

    for x in range(1,len(t)):

        t2.append(trad_to_int(t[x].value))

    print(t2)

    r=0

    for index in range(0,len(t2)):

       r+=float(t1[index])*float(t2[index])

    print(r)

if __name__ == '__main__':

    main()

跑脚本

NSSCTF{5030782.26}

ZIPBOMB

我也不知道他为啥没炸反正我直接解压了

010中得flag

NSSCTF{Z1p_B00m_d1sp0sal}

gif好像有点大

那确实挺大得最开始以为是隐写

看了看gif发现他只是长而已...

拆分gif脚本

from PIL import Image

import os

pic=Image.open("CTF.gif")

try:

    i=0

    while True:

        pic.seek(i)#搜寻文件帧数

        pic.save(str(i)+".png")

        i+=1

except:

    pass

然后有一帧里面有个二维码扫描即可

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