BUUCTF Crypto RSA4 wp

看到这一题，给到了三组三个n三个c，我们可以很容易的想到了中国剩余定理，但是我们发现这题并不是那么容易的，可以说是非常恶心人了，他题目给到的c和n并不是用的十进制，而是用的五进制，要先用int("*****",5)的代码转换为十进制才能计算。后来我们又发现，将题目通过CRT解出来并不对，看来并不是flag直接模的n，于是我们猜测e=3，果然，成功解出flag，脚本如下

// python2
import gmpy2
from functools import reduce
import sympy
    
def chinese_remainder(n, a):
    sum = 0
    prod = reduce(lambda a, b: a * b, n)

    for n_i, a_i in zip(n, a):
        p = prod // n_i
        sum += a_i * sympy.invert(p, n_i) * p
    return int(sum % prod)
def wu_shi(n):
    strlen=len(n)
    s=0
    for i in range(0,strlen):
        s+=int(n[i])*pow(5,strlen-i-1)
    return s

n1 = int(str(331310324212000030020214312244232222400142410423413104441140203003243002104333214202031202212403400220031202142322434104143104244241214204444443323000244130122022422310201104411044030113302323014101331214303223312402430402404413033243132101010422240133122211400434023222214231402403403200012221023341333340042343122302113410210110221233241303024431330001303404020104442443120130000334110042432010203401440404010003442001223042211442001413004),5)
c1 = int(str(310020004234033304244200421414413320341301002123030311202340222410301423440312412440240244110200112141140201224032402232131204213012303204422003300004011434102141321223311243242010014140422411342304322201241112402132203101131221223004022003120002110230023341143201404311340311134230140231412201333333142402423134333211302102413111111424430032440123340034044314223400401224111323000242234420441240411021023100222003123214343030122032301042243),5)

n2 = int(str(302240000040421410144422133334143140011011044322223144412002220243001141141114123223331331304421113021231204322233120121444434210041232214144413244434424302311222143224402302432102242132244032010020113224011121043232143221203424243134044314022212024343100042342002432331144300214212414033414120004344211330224020301223033334324244031204240122301242232011303211220044222411134403012132420311110302442344021122101224411230002203344140143044114),5)
c2 = int(str(112200203404013430330214124004404423210041321043000303233141423344144222343401042200334033203124030011440014210112103234440312134032123400444344144233020130110134042102220302002413321102022414130443041144240310121020100310104334204234412411424420321211112232031121330310333414423433343322024400121200333330432223421433344122023012440013041401423202210124024431040013414313121123433424113113414422043330422002314144111134142044333404112240344),5)

n3 = int(str(332200324410041111434222123043121331442103233332422341041340412034230003314420311333101344231212130200312041044324431141033004333110021013020140020011222012300020041342040004002220210223122111314112124333211132230332124022423141214031303144444134403024420111423244424030030003340213032121303213343020401304243330001314023030121034113334404440421242240113103203013341231330004332040302440011324004130324034323430143102401440130242321424020323),5)
c3 = int(str(10013444120141130322433204124002242224332334011124210012440241402342100410331131441303242011002101323040403311120421304422222200324402244243322422444414043342130111111330022213203030324422101133032212042042243101434342203204121042113212104212423330331134311311114143200011240002111312122234340003403312040401043021433112031334324322123304112340014030132021432101130211241134422413442312013042141212003102211300321404043012124332013240431242),5)


n=[n1,n2,n3]
c=[c1,c2,c3]
ans=chinese_remainder(n, c)
ans=gmpy2.iroot(ans,3)[0]
print hex(ans)[2:].decode('hex')
#noxCTF{D4mn_y0u_h4s74d_wh47_4_b100dy_b4s74rd!}