小明文攻击

小明文攻击的要求是e足够小

因为存在这个关系

m^e=c mod n
所以有
m^e=k*n + c  k为整数
m  = (k*n+c)^(1/e)

同时因为e足够小所以k足够小，可以通过穷举k来获得明文m

exp:

for e in range(1,5):#遍历e,已知e可以不用遍历
	for k in range(2000):#遍历k
		import Crypto.Util.number
			if __import__("gmpy2").iroot((k*n+flag),e)[1]:
				if b"NSSCTF" in Crypto.Util.number.long_to_bytes(__import__("gmpy2").iroot((k*n+flag),e)[0]):#条件判断
					print(Crypto.Util.number.long_to_bytes(__import__("gmpy2").iroot((k*n+flag),e)[0]))

[SWPUCTF 2021 新生赛]crypto5 | NSSCTF

from gmpy2 import *
from Crypto.Util.number import *
import sympy
import math
flag= 25166751653530941364839663846806543387720865339263370907985655775152187319464715737116599171477207047430065345882626259880756839094179627032623895330242655333
n= 134109481482703713214838023035418052567000870587160796935708584694132507394211363652420160931185332280406437290210512090663977634730864032370977407179731940068634536079284528020739988665713200815021342700369922518406968356455736393738946128013973643235228327971170711979683931964854563904980669850660628561419
for k in range(2000):
	import Crypto.Util.number
	for e in range(1,5):
		if __import__("gmpy2").iroot((k*n+flag),e)[1]:
			if b"NSSCTF" in Crypto.Util.number.long_to_bytes(__import__("gmpy2").iroot((k*n+flag),e)[0]):
				print(Crypto.Util.number.long_to_bytes(__import__("gmpy2").iroot((k*n+flag),e)[0]))