NCTF-2019-Crypto部分 复现

Sore

题目描述：
task.py

from string import ascii_letters
from flag import flag


ctoi = lambda x: ascii_letters.index(x) # 获得所有字母的字符串
itoc = lambda x: ascii_letters[x] # 将索引值转换为字母

key = flag.strip('NCTF{}')
len_key = len(key)

plaintext = open('plaintext.txt', 'r').read()

plain = ''.join(p for p in plaintext if p in ascii_letters)
cipher = ''.join( itoc( ( ctoi(p) + ctoi( key[i % len_key] ) ) % 52 )  for i,p in enumerate(plain) )

open('ciphertext.txt', 'w').write(cipher)
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ciphertext.txt:

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题目分析：
维吉尼亚密码解密

NCTF{vlbeunuozbpycklsjXlfpaq}

childRSA

题目描述：

from random import choice
from Crypto.Util.number import isPrime, sieve_base as primes
from flag import flag


def getPrime(bits):
    while True:
        n = 2
        while n.bit_length() < bits:
            n *= choice(primes)
        if isPrime(n + 1):
            return n + 1

e = 0x10001
m = int.from_bytes(flag.encode(), 'big')
p, q = [getPrime(2048) for _ in range(2)]
n = p * q
c = pow(m, e, n)

# n = 32849718197337581823002243717057659218502519004386996660885100592872201948834155543125924395614928962750579667346279456710633774501407292473006312537723894221717638059058796679686953564471994009285384798450493756900459225040360430847240975678450171551048783818642467506711424027848778367427338647282428667393241157151675410661015044633282064056800913282016363415202171926089293431012379261585078566301060173689328363696699811123592090204578098276704877408688525618732848817623879899628629300385790344366046641825507767709276622692835393219811283244303899850483748651722336996164724553364097066493953127153066970594638491950199605713033004684970381605908909693802373826516622872100822213645899846325022476318425889580091613323747640467299866189070780620292627043349618839126919699862580579994887507733838561768581933029077488033326056066378869170169389819542928899483936705521710423905128732013121538495096959944889076705471928490092476616709838980562233255542325528398956185421193665359897664110835645928646616337700617883946369110702443135980068553511927115723157704586595844927607636003501038871748639417378062348085980873502535098755568810971926925447913858894180171498580131088992227637341857123607600275137768132347158657063692388249513
# c = 26308018356739853895382240109968894175166731283702927002165268998773708335216338997058314157717147131083296551313334042509806229853341488461087009955203854253313827608275460592785607739091992591431080342664081962030557042784864074533380701014585315663218783130162376176094773010478159362434331787279303302718098735574605469803801873109982473258207444342330633191849040553550708886593340770753064322410889048135425025715982196600650740987076486540674090923181664281515197679745907830107684777248532278645343716263686014941081417914622724906314960249945105011301731247324601620886782967217339340393853616450077105125391982689986178342417223392217085276465471102737594719932347242482670320801063191869471318313514407997326350065187904154229557706351355052446027159972546737213451422978211055778164578782156428466626894026103053360431281644645515155471301826844754338802352846095293421718249819728205538534652212984831283642472071669494851823123552827380737798609829706225744376667082534026874483482483127491533474306552210039386256062116345785870668331513725792053302188276682550672663353937781055621860101624242216671635824311412793495965628876036344731733142759495348248970313655381407241457118743532311394697763283681852908564387282605279108
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题目分析：
Pollard’s p-1光滑

import gmpy2
from Crypto.Util.number import *
e = 0x10001
N = 32849718197337581823002243717057659218502519004386996660885100592872201948834155543125924395614928962750579667346279456710633774501407292473006312537723894221717638059058796679686953564471994009285384798450493756900459225040360430847240975678450171551048783818642467506711424027848778367427338647282428667393241157151675410661015044633282064056800913282016363415202171926089293431012379261585078566301060173689328363696699811123592090204578098276704877408688525618732848817623879899628629300385790344366046641825507767709276622692835393219811283244303899850483748651722336996164724553364097066493953127153066970594638491950199605713033004684970381605908909693802373826516622872100822213645899846325022476318425889580091613323747640467299866189070780620292627043349618839126919699862580579994887507733838561768581933029077488033326056066378869170169389819542928899483936705521710423905128732013121538495096959944889076705471928490092476616709838980562233255542325528398956185421193665359897664110835645928646616337700617883946369110702443135980068553511927115723157704586595844927607636003501038871748639417378062348085980873502535098755568810971926925447913858894180171498580131088992227637341857123607600275137768132347158657063692388249513
c = 26308018356739853895382240109968894175166731283702927002165268998773708335216338997058314157717147131083296551313334042509806229853341488461087009955203854253313827608275460592785607739091992591431080342664081962030557042784864074533380701014585315663218783130162376176094773010478159362434331787279303302718098735574605469803801873109982473258207444342330633191849040553550708886593340770753064322410889048135425025715982196600650740987076486540674090923181664281515197679745907830107684777248532278645343716263686014941081417914622724906314960249945105011301731247324601620886782967217339340393853616450077105125391982689986178342417223392217085276465471102737594719932347242482670320801063191869471318313514407997326350065187904154229557706351355052446027159972546737213451422978211055778164578782156428466626894026103053360431281644645515155471301826844754338802352846095293421718249819728205538534652212984831283642472071669494851823123552827380737798609829706225744376667082534026874483482483127491533474306552210039386256062116345785870668331513725792053302188276682550672663353937781055621860101624242216671635824311412793495965628876036344731733142759495348248970313655381407241457118743532311394697763283681852908564387282605279108

def Pollards_p_1(N):
    n = 2
    a = 2
    while True:
        a = pow(a,n,N)
        res = gmpy2.gcd(a-1,N)
        print(n)
        if res != 1 and res != N:
            print('n = ',n)
            print('p = ',res)
            return
        n += 1
# p = Pollards_p_1(N)
p = 178449493212694205742332078583256205058672290603652616240227340638730811945224947826121772642204629335108873832781921390308501763661154638696935732709724016546955977529088135995838497476350749621442719690722226913635772410880516639651363626821442456779009699333452616953193799328647446968707045304702547915799734431818800374360377292309248361548868909066895474518333089446581763425755389837072166970684877011663234978631869703859541876049132713490090720408351108387971577438951727337962368478059295446047962510687695047494480605473377173021467764495541590394732685140829152761532035790187269724703444386838656193674253139

q = N // p
phi = (p-1)*(q-1)
d = gmpy2.invert(e,phi)
m = pow(c,d,N)
print(long_to_bytes(m))

# NCTF{Th3r3_ar3_1ns3cure_RSA_m0duli_7hat_at_f1rst_gl4nce_appe4r_t0_be_s3cur3}
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easyRSA

题目描述：

from flag import flag

e = 0x1337
p = 199138677823743837339927520157607820029746574557746549094921488292877226509198315016018919385259781238148402833316033634968163276198999279327827901879426429664674358844084491830543271625147280950273934405879341438429171453002453838897458102128836690385604150324972907981960626767679153125735677417397078196059
q = 112213695905472142415221444515326532320352429478341683352811183503269676555434601229013679319423878238944956830244386653674413411658696751173844443394608246716053086226910581400528167848306119179879115809778793093611381764939789057524575349501163689452810148280625226541609383166347879832134495444706697124741
n = p * q

assert(flag.startswith('NCTF'))
m = int.from_bytes(flag.encode(), 'big')
assert(m.bit_length() > 1337)

c = pow(m, e, n)
print(c)
# 10562302690541901187975815594605242014385201583329309191736952454310803387032252007244962585846519762051885640856082157060593829013572592812958261432327975138581784360302599265408134332094134880789013207382277849503344042487389850373487656200657856862096900860792273206447552132458430989534820256156021128891296387414689693952047302604774923411425863612316726417214819110981605912408620996068520823370069362751149060142640529571400977787330956486849449005402750224992048562898004309319577192693315658275912449198365737965570035264841782399978307388920681068646219895287752359564029778568376881425070363592696751183359
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题目分析：
e与phi不互素，AMM算法求解

import random
import time
from tqdm import tqdm
from Crypto.Util.number import *
# About 3 seconds to run
def AMM(o, r, q):
    start = time.time()
    print('\n----------------------------------------------------------------------------------')
    print('Start to run Adleman-Manders-Miller Root Extraction Method')
    print('Try to find one {:#x}th root of {} modulo {}'.format(r, o, q))
    g = GF(q)
    o = g(o)
    p = g(random.randint(1, q))
    while p ^ ((q-1) // r) == 1:
        p = g(random.randint(1, q))
    print('[+] Find p:{}'.format(p))
    t = 0
    s = q - 1
    while s % r == 0:
        t += 1
        s = s // r
    print('[+] Find s:{}, t:{}'.format(s, t))
    k = 1
    while (k * s + 1) % r != 0:
        k += 1
    alp = (k * s + 1) // r
    print('[+] Find alp:{}'.format(alp))
    a = p ^ (r**(t-1) * s)
    b = o ^ (r*alp - 1)
    c = p ^ s
    h = 1
    for i in range(1, t):
        d = b ^ (r^(t-1-i))
        if d == 1:
            j = 0
        else:
            print('[+] Calculating DLP...')
            j = - discrete_log(d, a)
            print('[+] Finish DLP...')
        b = b * (c^r)^j
        h = h * c^j
        c = c^r
    result = o^alp * h
    end = time.time()
    print("Finished in {} seconds.".format(end - start))
    print('Find one solution: {}'.format(result))
    return result

def onemod(p,r): 
    t=random.randint(2,p)
    while pow(t,(p-1)//r,p)==1: 
         t=random.randint(2,p)
    return pow(t,(p-1)//r,p) 
 
def solution(p,root,e):  
    while True:
        g=onemod(p,e) 
        may=[] 
        for i in tqdm(range(e)): 
            may.append(root*pow(g,i,p)%p)
        if len(may) == len(set(may)):
            return may


def solve_in_subset(ep,p):
    cp = int(pow(c,inverse(int(e//ep),p-1),p))
    com_factors = []
    while GCD(ep,p-1) !=1:
        com_factors.append(GCD(ep,p-1))
        ep //= GCD(ep,p-1)
    com_factors.sort()

    cps = [cp]
    for factor in com_factors:
        mps = []
        for cp in cps:
            mp = AMM(cp, factor, p)
            mps += solution(p,mp,factor)
        cps = mps
    for each in cps:
        assert pow(each,e,p)==c%p
    return cps

e = 0x1337
p = 199138677823743837339927520157607820029746574557746549094921488292877226509198315016018919385259781238148402833316033634968163276198999279327827901879426429664674358844084491830543271625147280950273934405879341438429171453002453838897458102128836690385604150324972907981960626767679153125735677417397078196059
q = 112213695905472142415221444515326532320352429478341683352811183503269676555434601229013679319423878238944956830244386653674413411658696751173844443394608246716053086226910581400528167848306119179879115809778793093611381764939789057524575349501163689452810148280625226541609383166347879832134495444706697124741
n = p * q
c = 10562302690541901187975815594605242014385201583329309191736952454310803387032252007244962585846519762051885640856082157060593829013572592812958261432327975138581784360302599265408134332094134880789013207382277849503344042487389850373487656200657856862096900860792273206447552132458430989534820256156021128891296387414689693952047302604774923411425863612316726417214819110981605912408620996068520823370069362751149060142640529571400977787330956486849449005402750224992048562898004309319577192693315658275912449198365737965570035264841782399978307388920681068646219895287752359564029778568376881425070363592696751183359

m_p = solve_in_subset(e,p)
m_q = solve_in_subset(e,q)

for mpp in m_p:
    for mqq in m_q: 
        m = crt([int(mpp),int(mqq)],[p,q])
        flag = long_to_bytes(m)
        if b'NCTF{' in flag:
            print(flag)
            break

# b'NCTF{T4k31ng_Ox1337_r00t_1s_n0t_th4t_34sy}'
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babyRSA

题目描述：

from Crypto.Util.number import *
from flag import flag

def nextPrime(n):
    n += 2 if n & 1 else 1
    while not isPrime(n):
        n += 2
    return n

p = getPrime(1024)
q = nextPrime(p)
n = p * q
e = 0x10001
d = inverse(e, (p-1) * (q-1))
c = pow(bytes_to_long(flag.encode()), e, n)

# d = 19275778946037899718035455438175509175723911466127462154506916564101519923603308900331427601983476886255849200332374081996442976307058597390881168155862238533018621944733299208108185814179466844504468163200369996564265921022888670062554504758512453217434777820468049494313818291727050400752551716550403647148197148884408264686846693842118387217753516963449753809860354047619256787869400297858568139700396567519469825398575103885487624463424429913017729585620877168171603444111464692841379661112075123399343270610272287865200880398193573260848268633461983435015031227070217852728240847398084414687146397303110709214913
# c = 5382723168073828110696168558294206681757991149022777821127563301413483223874527233300721180839298617076705685041174247415826157096583055069337393987892262764211225227035880754417457056723909135525244957935906902665679777101130111392780237502928656225705262431431953003520093932924375902111280077255205118217436744112064069429678632923259898627997145803892753989255615273140300021040654505901442787810653626524305706316663169341797205752938755590056568986738227803487467274114398257187962140796551136220532809687606867385639367743705527511680719955380746377631156468689844150878381460560990755652899449340045313521804
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题目分析：

$$\begin{gathered} e\ast d-1=k\ast phi \\ d<phi\Rightarrow e>k \\ 爆破k得到phi \\ (p-1{)}^2<phi=(p-1)\ast (q-1)<(q-1{)}^2 \\ p,q相差很小，so可以在ph{i}^{1/2}上下浮动小范围内爆破出p-1或q-1 \\ 由此得到p,q，之后常规RSA得到flag \end{gathered}$$

from gmpy2 import *
from Crypto.Util.number import *
e = 0x10001
d = 19275778946037899718035455438175509175723911466127462154506916564101519923603308900331427601983476886255849200332374081996442976307058597390881168155862238533018621944733299208108185814179466844504468163200369996564265921022888670062554504758512453217434777820468049494313818291727050400752551716550403647148197148884408264686846693842118387217753516963449753809860354047619256787869400297858568139700396567519469825398575103885487624463424429913017729585620877168171603444111464692841379661112075123399343270610272287865200880398193573260848268633461983435015031227070217852728240847398084414687146397303110709214913
c = 5382723168073828110696168558294206681757991149022777821127563301413483223874527233300721180839298617076705685041174247415826157096583055069337393987892262764211225227035880754417457056723909135525244957935906902665679777101130111392780237502928656225705262431431953003520093932924375902111280077255205118217436744112064069429678632923259898627997145803892753989255615273140300021040654505901442787810653626524305706316663169341797205752938755590056568986738227803487467274114398257187962140796551136220532809687606867385639367743705527511680719955380746377631156468689844150878381460560990755652899449340045313521804
ed = e * d
for k in range(1,e):
    if (ed - 1) % k == 0:
        phi = (ed - 1) // k
        root = iroot(phi,2)[0]
        for p_1 in range(root - 2000,root + 2000):
            if phi % (p_1) == 0:
                p = p_1 + 1
                q = (phi // p_1) + 1
                m = long_to_bytes(pow(c,d,p * q))
                if b'NCTF' in m:
                    print(m)
                    break
# NCTF{70u2_nn47h_14_v3ry_gOO0000000d}
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考点：维吉尼亚，Pollard’s p-1光滑，AMM，(知e,d,c，求p,q)