[安洵杯 2019]JustBase

VGhlIGdlb@xvZ#kgb@YgdGhlIEVhcnRoJ#Mgc#VyZmFjZSBpcyBkb@!pbmF)ZWQgYnkgdGhlIHBhcnRpY#VsYXIgcHJvcGVydGllcyBvZiB#YXRlci$gUHJlc@VudCBvbiBFYXJ)aCBpbiBzb@xpZCwgbGlxdWlkLCBhbmQgZ@FzZW(!cyBzdGF)ZXMsIHdhdGVyIGlzIGV$Y@VwdGlvbmFsbHkgcmVhY#RpdmUuIEl)IGRpc#NvbHZlcywgdHJhbnNwb#J)cywgYW%kIHByZWNpcGl)YXRlcyBtYW%%IGNoZW!pY@FsIGNvbXBvdW%kcyBhbmQgaXMgY@(uc#RhbnRseSBtb@RpZnlpbmcgdGhlIGZhY@Ugb@YgdGhlIEVhcnRoLiBFdmFwb#JhdGVkIGZyb@)gdGhlIG(jZWFucywgd@F)ZXIgdmFwb#IgZm(ybXMgY@xvdWRzLCBzb@!lIG(mIHdoaWNoIGFyZSB)cmFuc#BvcnRlZCBieSB#aW%kIG(@ZXIgdGhlIGNvbnRpbmVudHMuIENvbmRlbnNhdGlvbiBmcm(tIHRoZSBjbG(!ZHMgcHJvdmlkZXMgdGhlIGVzc@VudGlhbCBhZ@VudCBvZiBjb@%)aW%lbnRhbCBlcm(zaW(uOiByYWluLlRoZSByYXRlIGF)IHdoaWNoIGEgbW(sZWN!bGUgb@Ygd@F)ZXIgcGFzc@VzIHRob#VnaCB)aGUgY#ljbGUgaXMgbm()IHJhbmRvbQpBbmQgdGhlIGZsYWcgaXM^IENURnsyMi!RV)VSVFlVSU*tUExLSkhHRkRTLUFaWENWQk%NfQ==

把)!@#$%^&*(替换成0123456789

这里我们可以使用记事本替换，也可以使用python脚本

import base64
value = 'VGhlIGdlb@xvZ#kgb@YgdGhlIEVhcnRoJ#Mgc#VyZmFjZSBpcyBkb@!pbmF)ZWQgYnkgdGhlIHBhcnRpY#VsYXIgcHJvcGVydGllcyBvZiB#YXRlci$gUHJlc@VudCBvbiBFYXJ)aCBpbiBzb@xpZCwgbGlxdWlkLCBhbmQgZ@FzZW(!cyBzdGF)ZXMsIHdhdGVyIGlzIGV$Y@VwdGlvbmFsbHkgcmVhY#RpdmUuIEl)IGRpc#NvbHZlcywgdHJhbnNwb#J)cywgYW%kIHByZWNpcGl)YXRlcyBtYW%%IGNoZW!pY@FsIGNvbXBvdW%kcyBhbmQgaXMgY@(uc#RhbnRseSBtb@RpZnlpbmcgdGhlIGZhY@Ugb@YgdGhlIEVhcnRoLiBFdmFwb#JhdGVkIGZyb@)gdGhlIG(jZWFucywgd@F)ZXIgdmFwb#IgZm(ybXMgY@xvdWRzLCBzb@!lIG(mIHdoaWNoIGFyZSB)cmFuc#BvcnRlZCBieSB#aW%kIG(@ZXIgdGhlIGNvbnRpbmVudHMuIENvbmRlbnNhdGlvbiBmcm(tIHRoZSBjbG(!ZHMgcHJvdmlkZXMgdGhlIGVzc@VudGlhbCBhZ@VudCBvZiBjb@%)aW%lbnRhbCBlcm(zaW(uOiByYWluLlRoZSByYXRlIGF)IHdoaWNoIGEgbW(sZWN!bGUgb@Ygd@F)ZXIgcGFzc@VzIHRob#VnaCB)aGUgY#ljbGUgaXMgbm()IHJhbmRvbQpBbmQgdGhlIGZsYWcgaXM^IENURnsyMi!RV)VSVFlVSU*tUExLSkhHRkRTLUFaWENWQk%NfQ=='
reg = ')!@#$%^&*('
for i in value
	for j in range(10):
		if i == reg[j]
			m = m.replace(i,str(j))
printf(base64.b64encode(m))

[SWPUCTF 2021 新生赛]traditional

字符替换解密

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{}

c =  "震坤艮 震艮震 坤巽坤 坤巽震 震巽兑 震艮震 震离艮 震离艮"
table={'乾':'111','坤':'000','震':'001','兑':'011','离':'101','艮':'100','巽':'110'}
t = ""
for i in c:
	if i==' ':
		t+=" " //这步也是很关键的一点
	else:
		t += table[i]
flag=""
for i in t.split(' '):
	flag+=chr(int(i,2))
    //int(i,2)是把i当作二进制数转成整型十进制然后再将十进制数转为
    //字符串
print('NSSCTF{%s}'%(flag))

还需要注意两点

[BJDCTF 2020]base??

字符替换 --》 base64解密

下载附件让我来康康

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: '='}

chipertext:
FlZNfnF6Qol6e9w17WwQQoGYBQCgIkGTa9w3IQKw

脚本

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])


//让我们来分析一下，先是用for语句把作者的这个交换表先做出来 key1
//然后写出标准的64位字符表  key2
//用b来存储密文
//关键的一点就在于下面这个for循环
	for i in enc
	//逐字把enc中的字符分离出来。i的值就是每次分离出来的值。
	//key1.find(i)就是在交换表中找出i的值所在的下标，然后
	//赋值给a
	//然后将下标在标准表所对应的值找出来
	最后拼接到一起

然后进行base64解码，拿到flag