BUUCTF--Dangeous RSA（小e）

对于e很小，可以直接采取爆破的手段，直接上代码

点击查看代码

#python3
## -*- coding: utf-8 -*-#
import binascii

from gmpy2 import iroot
import libnum
e = 0x3 #e很小
n = 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
c = 0x10652cdfaa6b63f6d7bd1109da08181e500e5643f5b240a9024bfa84d5f2cac9310562978347bb232d63e7289283871efab83d84ff5a7b64a94a79d34cfbd4ef121723ba1f663e514f83f6f01492b4e13e1bb4296d96ea5a353d3bf2edd2f449c03c4a3e995237985a596908adc741f32365

k = 0
while 1:
    res = iroot(c+k*n,e)  #c+k*n 开3次方根 能开3次方即可，对x开n次方，返回整数和布尔值（能被完全开方为整数，则返回True；不能，则返回False）
    #print(res)
    #res = (mpz(13040004482819713819817340524563023159919305047824600478799740488797710355579494486728991357), True)
    if(res[1] == True):
        print(libnum.n2s(int(res[0]))) #转为字符串,十进制to字符串
        break
    k=k+1
#b'flag{25df8caf006ee5db94d48144c33b2c3b}'