这么简单的题目搞了我那么久，森森感觉自己菜的不行。。。。哎，努力吧少年，BXS已经全国第二了。
嗯，废话不说，这道题目来自实验吧的“warmup“，附上链接 http://www.shiyanbar.com/ctf/73
开始下载出来了两个文件。一个是.pub还有一个是TXT，打开后发现了什么Public Key，然后隐隐约约记得曾经在哪里见过类似的文件，后来发现这是用“openssl”加密过的一段密码，然后把这个东西.pub文件扔到了kali下，直接用命令去跑（这里附上命令 openssl rsa -pubin -text -modulus -in warmup.pub ），然后出来了一个这
Public-Key: (2050 bit)
Modulus:
03:67:19:8d:6b:56:14:e9:58:13:ad:d8:f2:2a:47:
17:bc:72:be:1e:ab:d9:33:d1:b8:69:44:fd:b7:5b:
8e:d2:30:be:62:d7:d1:b6:9d:22:20:95:c1:28:c8:
6f:82:01:2e:cb:11:61:91:fd:9d:01:8a:6d:02:f8:
4d:b2:7b:c5:1a:21:30:7d:c8:6f:4b:f7:71:c6:91:
c1:43:e5:ab:e5:49:b5:bd:2d:6e:b1:a2:1f:d6:27:
0e:7e:1b:48:fe:06:11:fb:b2:e1:b0:b3:52:4e:6f:
4d:e8:b4:e4:a3:45:da:44:a1:3d:e8:25:b7:26:08:
db:6c:7c:4a:40:b7:82:66:e6:c8:7b:bf:de:f6:b4:
83:81:d4:9c:45:07:a5:8b:cd:47:b7:6d:64:b4:59:
08:b1:58:bd:7e:bc:4d:ac:b0:b1:cf:d6:c2:c1:95:
74:f4:0e:b2:ef:d0:e9:e1:0d:c7:00:5c:ad:39:bc:
af:52:b9:ea:c3:87:33:68:d6:90:31:c5:e7:24:68:
4a:44:f0:68:ef:d1:d3:dc:09:6d:9b:5d:64:11:e5:
8b:de:e4:3e:46:b9:9a:0d:04:94:b9:db:28:19:5a:
f9:01:af:f1:30:d4:a6:e2:03:da:d0:8d:a5:7f:a7:
e4:02:62:a5:ba:db:2a:32:3e:da:28:b4:46:96:ab:
30:5d
Exponent:
00:f3:95:9d:97:8e:02:eb:9f:06:de:f3:f3:35:d8:
f8:af:d7:60:99:51:dd:ac:60:b7:14:b6:c2:2a:f0:
fa:91:2f:21:0b:34:20:6b:d2:4a:96:01:c7:8d:f4:
a0:27:5f:10:7f:d3:ab:55:2d:95:05:7e:b9:34:e7:
1b:dd:cd:70:45:c2:4b:18:58:7b:8c:8f:cf:5a:dd:
4c:5d:83:f0:c7:7c:94:dc:9c:50:cb:e4:38:e2:b6:
7b:af:d3:16:33:b6:aa:f1:78:1d:90:c3:ad:6f:03:
d0:37:b3:32:18:01:b2:35:46:d4:83:e6:7e:26:06:
7f:7b:22:34:7d:db:c0:c2:d5:92:ce:81:4c:bf:5d:
fc:cc:14:14:37:f1:4e:0b:39:90:f8:80:61:e5:f0:
ba:e5:f0:1e:3f:a7:0d:b0:e9:60:5e:7c:fd:57:5e:
9c:81:ef:ee:c5:29:c3:3f:d9:03:7a:20:fd:8a:cd:
51:3a:c9:63:77:68:31:3e:63:f9:83:8a:e3:51:1c:
dd:0a:9a:2b:51:6f:21:48:c8:d4:75:a3:60:a0:63:
59:44:97:39:ee:cd:25:1a:bb:42:b0:14:57:3e:43:
9f:2f:a4:57:35:57:b2:56:99:ff:c1:1e:63:1c:e8:
ee:97:5a:86:e7:e2:72:bc:f5:f7:6a:93:45:03:48:
fe:3f
Modulus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
writing RSA key
-----BEGIN PUBLIC KEY-----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-----END PUBLIC KEY-----
就是这个文件，我找了老半天都没有分清E和N到底Modulu哪个是哪个，，还是太嫩了。后天没办法，我就各种百度也没找到公钥E在哪，后来问表哥，他说那个Exponent16进制就是，天真的我还想捷径去找如何输入规范，后来发现没辙。所以我用txt文件把冒号和空格去掉。勉强得到了
N:
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
E:
f3959d978e02eb9f06def3f335d8f8afd7609951ddac60b714b6c22af0fa912f210b34206bd24a9601c78df4a0275f107fd3ab552d95057eb934e71bddcd7045c24b18587b8c8fcf5add4c5d83f0c77c94dc9c50cbe438e2b67bafd31633b6aaf1781d90c3ad6f03d037b3321801b23546d483e67e26067f7b22347ddbc0c2d592ce814cbf5dfccc141437f14e0b3990f88061e5f0bae5f01e3fa70db0e9605e7cfd575e9c81efeec529c33fd9037a20fd8acd513ac9637768313e63f9838ae3511cdd0a9a2b516f2148c8d475a360a06359449739eecd251abb42b014573e439f2fa4573557b25699ffc11e631ce8ee975a86e7e272bcf5f76a93450348fe3f
之后就方便了。现在就是求D的时刻。然而，天真的我跑了一下N的长度。。呵呵呵，这个N有618位，这种大素数的分解就成了大问题，后天我在网上查，终于找到了方法：“用git上的一个py文件——rsa-wiener-attack-master”，然后看他的代码，然后放进去跑了一下，果然跑出来了（惊喜脸0~0），得到私钥
D：4221909016509078129201801236879446760697885220928506696150646938237440992746683409881141451831939190609743447676525325543963362353923989076199470515758399
之后就是套路解RSA了 ：代码附上：然后解出来M，然后用binascii这个库把得出来的十进制转换为字符串，得出结果。。23333333333333333
终于写完，一会看看椭圆加密。开心的一天，继续努力！！！
import gmpy2
c=gmpy2.mpz(60627434129028814567404969251697807879433775099089276036015410639261901886263220690479587955645647729237126627894599166136756811335873604339498869129719440656776867314172756225023754233476201217843405147707204606506489409981939998604644568471901388086489470619171927867919622384085722093956759241836499067639379179411120052281316958208007504576202137300342827418254852620085360302233358835951034932808393567658291851385064158525774666702382955852511233053362902277921857534414199997716231020668999139795251740005513044276574836799364555665560793535452231662835543599932725511766192196706414172508995359134072584232582)
d=gmpy2.mpz(4221909016509078129201801236879446760697885220928506696150646938237440992746683409881141451831939190609743447676525325543963362353923989076199470515758399)
N=gmpy2.mpz(109966163992903243770643456296093759130737510333736483352345488643432614201030629970207047930115652268531222079508230987041869779760776072105738457123387124961036111210544028669181361694095594938869077306417325203381820822917059651429857093388618818437282624857927551285811542685269229705594166370426152128895901914709902037365652575730201897361139518816164746228733410283595236405985958414491372301878718635708605256444921222945267625853091126691358833453283744166617463257821375566155675868452032401961727814314481343467702299949407935602389342183536222842556906657001984320973035314726867840698884052182976760066141)
message = pow(c,d,N)
print message
import binascii
m=344057191085314093305077619578446205
mm=hex(m).rstrip("L")
print mm[2:]
print binascii.unhexlify(mm[2:])
————————————Made By Pinging
