Problem: [柏鹭杯 2021]试试大数据分解?
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思路
- 解题大致思路
EXP
```from Crypto.Util.number import *
import base64
n = 980710843783426929436352108445477207723530613536846588304753445659989738992892689598648497
e = 99991
p = 948539437740472240970258995719507356652939947
q = 1033916782753483187367063564275935620987269651
c1 = open('C://Users//HP//Downloads//C1_RSA//flag.enc1','rb').read()
c1 = bytes_to_long(base64.b64decode(c1))
phi = (p-1)*(q-1)
d = inverse(e,phi)
m = pow(c1,d,n)
print(m)
print(long_to_bytes(m))
print((long_to_bytes(0x00666c61677b495345432d49)))
c2 = open('C://Users//HP//Downloads//C1_RSA//flag.enc2','rb').read()
c2 = bytes_to_long(base64.b64decode(c2))
phi = (p-1)*(q-1)
d = inverse(e,phi)
m = pow(c2,d,n)
print(m)
print(long_to_bytes(m))
print((long_to_bytes(0x007235574d5f473441666276)))
c3 = open('C://Users//HP//Downloads//C1_RSA//flag.enc3','rb').read()
c3 = bytes_to_long(base64.b64decode(c3))
phi = (p-1)*(q-1)
d = inverse(e,phi)
m = pow(c3,d,n)
print(m)
print(long_to_bytes(m))
print((long_to_bytes(0x00785f6d534d5f556766387a)))
c4 = open('C://Users//HP//Downloads//C1_RSA//flag.enc4','rb').read()
c4 = bytes_to_long(base64.b64decode(c4))
phi = (p-1)*(q-1)
d = inverse(e,phi)
m = pow(c4,d,n)
print(m)
print(long_to_bytes(m))
print((long_to_bytes(0x0052416f4d6b594350787d)))
print(long_to_bytes(0x00666c61677b495345432d49)+long_to_bytes(0x007235574d5f473441666276)+long_to_bytes(0x00785f6d534d5f556766387a)+long_to_bytes(0x0052416f4d6b594350787d))
## 总结
* 对该题的考点总结