每天的CTF小练--6.5(ascll码高级运用)

题目：[HUBUCTF 2022 新生赛]baby_encrypt

hint：

781612443113954655886887407898899451044114412011257135914071455155316031651170318041861191719652013207021272183228423832485254125932643269827992924

注意查看前面的数字，这题不想现代密码，

chr(78)
chr(161-78)
chr(244-161)
chr(311-244)
chr(395-311)
chr(465-395)###
N
S
S
C
T
F

就是将flag转化为前面的ascll码数字相加

这题反映了不一定要多强，主要你不放弃，就有可能做出来，

特别注意数字转ascll码

写代码直接解：

整理一下

78,161,244,311,395,465,588,688,740,789,889,945,1044,1144,1201,1257,1359,1407,1455,1553,1603,1651,1703,1804,1861,1917,1965,2013,2070,2127,2183,2284,2383,2485,2541,2593,2643,2698,2799,2924

s=[0,78,161,244,311,395,465,588,688,740,789,889,945,1044,1144,1201,1257,1359,1407,1455,1553,1603,1651,1703,1804,1861,1917,1965,2013,2070,2127,2183,2284,2383,2485,2541,2593,2643,2698,2799,2924]
flag=""
for i in range(0,len(s)-1):c=chr(s[i+1]-s[i])flag=flag+c
print(flag)#NSSCTF{d41d8cd98f00b204e9800998ecf8427e}

注意数组里面第一个位置0的运用，没有0没有N，这是数组索引的关系，第一个位置为0