[SWPUCTF 2021 新生赛]gif好像有点大

帧解一下

找到这个二维码用软件CQR解开一下

得到flag

NSSCTF{The_G1F_ls_T00_b1g}

[BJDCTF 2020]base??

给了我们base64加密的密文

用python直接解密

import base64
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: '='}
key1=''
#字典的键和键值反过来才是我们想用的字典，直接把它转化为字符串好点
for i in dict:
    key1+=dict[i]

key2='ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/='#标准表
enc='FlZNfnF6Qol6e9w17WwQQoGYBQCgIkGTa9w3IQKw'#密文
b=''
for i in enc:
   a=key1.find(i) #寻找密文在换表中的下标
   b+=key2[a]  #转换为标准表的密文
#解base
flag=base64.b64decode(b) 
print("NSSCTF"+str(flag)[5:100])

NSSCTF{D0_Y0u_kNoW_Th1s_b4se_map}

这是一个GET传参，使用了file伪协议

直接输入伪协议找flag就行

?file=php://filter/read=convert.base64-encode/resource=flag.php

给了我们一串密文

用随波逐流解密一下

NSSCTF{425cd42f-fba6-4be2-9824-55c116a7cf04}