baigeiRSA 1.首先呢,下载附件.◡.
附件给出了以下信息
n = 88503001447845031603457048661635807319447136634748350130947825183012205093541
c = 40876621398366534035989065383910105526025410999058860023908252093679681817257import libnum
from Crypto.Util import n…
题目:
from Crypto.Util.number import *
from secret import flagm bytes_to_long(flag)
n 1
for i in range(15):n *getPrime(32)
e 65537
c pow(m,e,n)
print(fn {n})
print(fc {c})n 152412082177688498871800101395902107678314310182046454156816957…