2019科成安洵杯WP

ez

搜索flag就出来了

小游戏

以前一道题改的，直接定位关键函数

Code

box1=[18,64,98,5,2,4,6,3,6,48,49,65,32,12,48,65,31,78,62,32,49,32,1,57,96,3,21,9,4,62,3,5,4,1,2,3,44,65,78,32,16,97,54,16,44,52,32,64,89,45,32,65,15,34,18,16,0]

box2=[123,32,18,98,119,108,65,41,124,80,125,38,124,111,74,49,83,108,94,108,84,6,96,83,44,121,104,110,32,95,117,101,99,123,127,119,96,48,107,71,92,29,81,107,90,85,64,12,43,76,86,13,114,1,117,126,0]
flag=list(range(57))

for i in range(57):
    flag[i]=box1[i]^box2[i]^0x14
    print(chr(flag[i]),end="")  

# flag{}tdsa|S>ntXsHwndX6tXq2u~Xnis0ubtsni`Xe1sXhsobutXfubXi7sz}

我要vip

Android Killer修改switch强行进入vip函数即可

goto :pswitch_1

flag{czADA_SA_ddad_aijdA}

不知道什么算法

触法异常进入success函数，看到div可以确定是除0异常

40135A除零异常
输入必须等于pop eax = 401353

flag{401353}

一个简单的cm

Upack壳，脱壳机脱不干净，esp定律完事

逻辑比较简单，逆回去即可

Code

# OEP 401ffc

str1 = list("UESTCDCTF2019JustForhappy")

box = list("abcdefghiABCDEFGHIJKLMNjklmn0123456789opqrstuvwxyzOPQRSTUVWXYZ")

key = list(range(len(str1)))

for i in range(len(str1)):
	key[i]=box.index(str1[i])
	print(key[i],end=',')

k=[]
k=key

for i in range(len(key)):
	if k[i]>=0 and k[i]<=9:
		k[i]=k[i]+48
	elif k[i]>=36 and k[i]<=64:
		k[i]=k[i]+29
	elif k[i]>=10 and k[i]<=35:
		k[i]=k[i]+87
	print("flag{",chr(key[i]),"}",end="")