I hope the epidemic will end soon
Note: the * is played after the game
Always wp wait for the legend of Nanshen to post the team wp link
Misc
soEasyCheckin
base32, but there is a problem. The countdown appears 0 $, 0 – > O, $- > s, and a string of hex is obtained.
As a result, there was another problem with hex, specifically 83 ¥ 6988ee here
According to the rule, the first byte of every 6 bytes is e, and then ¥ is replaced by E
Get the code of socialist core values, but there is still one place that is wrong. There is a paragraph in the middle: harmony dies Mingping
Then change his death casually. I change it to "rich and strong"
Decoded SET
According to the Pinyin, you can know that it is Yuan2
So the final flag is:
SET
Defeat the virus
After opening the game, I found that there was no response after playing the end shadow dragon. The flag was hidden in the end poem, so I went directly to find the text
Under. minecraft/version / Longyuan war "epidemic". jar
Change it to zip, find assets/minecraft/texts/end.txt, and get 11f9sacbbbwkticlydtnf2yiefthxdfigpxf
base62 decoding is enough
SETCTF
SOS
Dial tone, step on the correct to combine the flag
Record with your mobile phone, then convert m4a format to wav format, and then DTMF. Because there are always problems with recording, you have recorded it three times in total
There is no 8 in the front. This time, there is finally 8, so it should be 6830AB1C75. Get the flag
EasySteg
Brother, don't set the ball. If you put it down again, the God cried
JK is a single graph blind watermark (java), and imagein can also be used
Then there is a ciphertext composed of a string of tab and space in the annotation of flag.rar, which turns space to 0 and tab to 1
Then unzip it. There is another PNG at the end of flag.png. That PNG obviously has the word flag. You can see it by reversing the color with stegsolve
flag. Combined with 01 string, it is obtained as Huffman coding, which can be decoded by using the script of hongminggu
The correspondence can be understood by looking at the script
import copy import re def dfs(c, d): if len(c.keys()) == 1: # g = {'j':29,'z':31,'7':25,'e':31,'l':23,'6':37,'4':32,'p':38,'h':27,'g':26,'x':28,'i':25,'u':27,'n':25,'8':36,'0':24,'o':23,'c':28,'y':24,'1':29,'b':26,'m':27,'2':28,'v':25,'d':33,'f':28,'9':33,'t':21,'w':22,'a':31,'r':24,'s':16,'k':32,'5':25,'q':23,'3':32,'{':1,'-':4,'}':1,} # num = 0 # for k in g.keys(): # num += g[k] * len(d[k]) # print(num) # print(c, d) g = {} for k in d.keys(): g[d[k]] = k a = '1110111111000100001111000011010001101111100110100011110111100010100111110111001101111100011000111111111101011100011111100111000011010001111010011011111001110111100010000111001110011110111000111111100111011111011011101011110111110101101101101000110101011101011111111011110111101101110101111010011010110100011100100011010101111010111110110110111100011010111010111110110110111001001011011110100111010111111010111111011110111101001111010111110111101011100100111100101011011101110111111001010110100100110111110011111001101000111110111001011001110000111000011110000110111110011000011100001101100110100011100001111110011110000110110011010001110011101100001110110001001111111110010110011010001111011110110010011011111000001111100010010001001111101110110111111101101001001001011011101111101101101111001101001011011111100111110110110110110111110111010110111011110011111011011011010000110111111001111110111100111010111110011010011011110101001101110111100110110111101001101011101011111001101111011101101010111010111111001111110111010110110110101001101011111101001101000111010010111000010011001110111110111101101111101101110111010110100101101101101011010111111101111001110111111110011010110100101101111011011101100111001000001111001100100101011000000111100101011001000101100100000011001011001001000001111001100100101011000000111100101011001011111111001010001010001010010000011110011001001010110000001111001010110010001011001010001010000111100100000111100110010010101100000011110010101100101010010101100101010010000011110011001001010110000001111001010110010011001001010110000011011111001000001111001100100101011000000111100101000101000110111110010000001100101010010000011110011001001010010101100000011110010101100100010001001000011111110001010010000011110011001001010110000001111001010110010111111110010100010110001000100100000111100110010010101100000011110010100010100011011111001000000110010101001000001111001100100101011000000111100101011001001100100101011000001101111100100000111100110010010101100000011110010101100100010110010000001100101100100100000111100110010010101100000011110010100010100010100101011000001101111100100000111100110010010101100000011110010100101011001000101100100000011001011001001000001111001100100101011000000111100101011001011111111001010001010001010010000011110011001001010110000001111001010110010001000100100001111111000101001000001111001100100101011000000111100101011001000101100100000011001011001001000001111001100100101011000000111100101011001010100100000011001011111111001000001111001100100101011000000111100101011001000101100100000011001011001001000001111001100100101011000000111100101011001000100010010100100001111111000101001000001111001100100101011000000111100101011001010100100000011001011111111001000001111001100100101011000000111100101000101000101001010110000011011111001000001111001100100101011000000111100101000101000110111110010000001100000011110010000011110011001001010110000001111001010001010000001001010001010000111111011010011110011101011111011100001011010110101111010001111100110111110101111110110101010011101111101100100000101111011010111101101101110011001110001100011111001110010000010111100010111101111111101101101110000111010000101111110001100000110001110010100100000110000101110000100010110111100000101110011111110000111011010111101001101001101111101000100101111101101011111100111111001110001100001000011011111000001111100010011001110111101111111011111001000111000011101101110000111011011110011101111111011011010110100011111000100101111111011100000100000101110010111100011010110100011111101111001110101111010111011011100100110011101111100111101111100000001001001101001101111111101101011110111011110011101001101111010100110111011101011110011100000011010010111111000100110011101111101011110111110111' m = '' st = 0 while st < len(a): ed = st + 1 while ed <= len(a): if a[st:ed] in g.keys(): m += g[a[st:ed]] break else: ed += 1 st = ed print(m) else: k0 = list(c.keys())[0] k1 = list(c.keys())[1] if c[k0] > c[k1]: k0, k1 = k1, k0 for k in list(c.keys())[2:]: if c[k] < c[k1]: if c[k] < c[k0]: k0, k1 = k, k0 else: k1 = k for a in k0: d[a] = '0' + d[a] for a in k1: d[a] = '1' + d[a] c[k0+k1] = c[k0]+c[k1] del c[k0] del c[k1] dfs(copy.deepcopy(c), copy.deepcopy(d)) c = {'n':9,'S':2,'B':3,'I':6,'U':6,'L':1,'O':3,'R':2,'F':3,'Y':1,'G':2,'H':3,'v':9,'N':2,'M':1,'Z':1,'D':6,'T':1,'h':18,'o':16,'u':17,'b':9,'s':7,'r':4,'g':8,'f':10,'a':14,'m':2,'i':20,'p':2,'e':14,'w':7,'q':2,'y':8,'P':1,'J':2,'E':2,'C':1,'V':1,'A':1,'j':4,'k':5,'x':5,'t':3,'c':6,'8':24,'9':56,'{':1,'3':217,'d':97,'4':83,'6':54,'0':25,'1':12,'2':10,'5':8,'7':10,'}':1,'l':4,'Q':1,'W':1,'z':1} d = {} for k in c.keys(): d[k] = '' dfs(copy.deepcopy(c), copy.deepcopy(d)) #nSBIULORFYGHvNMIOUZSDTNhnoubuosrgfbouvasmruiohauiopewgvfbwuivpqynqwUPIFJDDBUEDUIDHBUIDCVHJIOAejikxneiwkyiohwehiooiuyhiosehfhuiaetyhovauieyrghfuotgvac89xcboiyuweagihniaweohuilagsieufrcb78QWEGF678Rniolsdf149687189735489246avaeukighf6497ejixcnbmlolohnbasik2647893hasfhuvzxchbjkaefgyhuetyuhjadfxcvbn
Take the content in {} to hex and find that it is Quoted Printable, but there are two times = =, so delete it repeatedly and get it
😩💌😏📓🐞🎣🗼💇🎣🐞😩🌞😩💌🗼😩📢😩🗼📢🌞🎯🍏
It's not base100, so take it to try a foreign ctfshow moon cake cup used to be made by la guy_ Used by Gong Chanjuan
https://github.com/pavelvodrazka/ctf-writeups/tree/master/hackyeaster2018/challenges/egg17
Splice them to get a flag
flag
*ez_misc
hint is zero width. It refers to an esolang very similar to piet, and then combined with ctfshowbabylsbwithhelicopter of the eight gods in June
You know, it's brainloller
Then open 010, and it will be prompted that it is the password of steghide behind, and the CRC will report an error. Once the CRC script explodes, it will be found that the correct width is 14 and the height is 12
Solve with bftools, bftools.exe decode brainroller bf.png, and then decode the BF to get the password Hello Worl
But I couldn't solve it, so I used the script I wrote in June to solve it https://blog.csdn.net/qq_42880719/article/details/117479024
It's Hello Worle!, I guess Hello World!
Combined with the hint on the new topic, I have tried to get the password
Hello Worl Hello_Worl Hello Worle! Hello_Worle! Hello World! Hello_World! Hello World Hello_World
Unfortunately, none of them are right. The author's friend asked the author, and his friend also said that he could not solve it, but the author can solve it (it seems that it uses local accessories)
So I always feel that it's the problem of the attachment of the competition topic?
- 17:02 -: after an hour and a half of positive feedback
Just bftools.exe, replace the space with the underscore Hello_Worl solves steghide and gets a text
Then a bear said, and then it was over
Re
EasyRe
flag is in the constant..
flag