第十三届全国大学生信息安全竞赛 WriteUp

国赛 WP

0x01 签到

flag{同舟共济扬帆起，乘风破浪万里航。}

0x02 the_best_ctf_game

拖入winhex中查看

注意观察得到flag！

0x03 z3

无壳，拖入IDA中查看，反编译得到代码！

分析得到，Dst为int型，在&unk_404020查看数据

编写长长的脚本如下：

from z3 import *
v4 = [0x4f17,0x9cf6,0x8ddb,0x8ea6,0x6929,0x9911,0x40a2,0x2f3e,0x62b6,0x4b82,0x486c,0x4002,0x52d7,0x2def,
0x28dc,0x640d,0x528f,0x613b,0x4781,0x6b17,0x3237,0x2a93,0x615f,0x50be,0x598e,0x4656,0x5b31,0x313a,0x3010,
0x67fe,0x4d5f,0x58db,0x3799,0x60a0,0x2750,0x3759,0x8953,0x7122,0x81f9,0x5524,0x8971,0x3a1d]
print(len(v4))
v46 = Int('v46')
v47 = Int('v47')
v48 = Int('v48')
v49 = Int('v49')
v50 = Int('v50')
v51 = Int('v51')
v52 = Int('v52')
v53 = Int('v53')
v54 = Int('v54')
v55 = Int('v55')
v56 = Int('v56')
v57 = Int('v57')
v58 = Int('v58')
v59 = Int('v59')
v60 = Int('v60')
v61 = Int('v61')
v62 = Int('v62')
v63 = Int('v63')
v64 = Int('v64')
v65 = Int('v65')
v66 = Int('v66')
v67 = Int('v67')
v68 = Int('v68')
v69 = Int('v69')
v70 = Int('v70')
v71 = Int('v71')
v72 = Int('v72')
v73 = Int('v73')
v74 = Int('v74')
v75 = Int('v75')
v76 = Int('v76')
v77 = Int('v77')
v78 = Int('v78')
v79 = Int('v79')
v80 = Int('v80')
v81 = Int('v81')
v82 = Int('v82')
v83 = Int('v83')
v84 = Int('v84')
v85 = Int('v85')
v86 = Int('v86')
v87 = Int('v87')
s = Solver()
s.add(v4[0] == 34 * v49 + 12 * v46 + 53 * v47 + 6 * v48 + 58 * v50 + 36 * v51 + v52)
s.add(v4[1] == 27 * v50 + 73 * v49 + 12 * v48 + 83 * v46 + 85 * v47 + 96 * v51 + 52 * v52)
s.add(v4[2] == 24 * v48 + 78 * v46 + 53 * v47 + 36 * v49 + 86 * v50 + 25 * v51 + 46 * v52)
s.add(v4[3] == 78 * v47 + 39 * v46 + 52 * v48 + 9 * v49 + 62 * v50 + 37 * v51 + 84 * v52)
s.add(v4[4] == 48 * v50 + 14 * v48 + 23 * v46 + 6 * v47 + 74 * v49 + 12 * v51 + 83 * v52)
s.add(v4[5] == 15 * v51 + 48 * v50 + 92 * v48 + 85 * v47 + 27 * v46 + 42 * v49 + 72 * v52)
s.add(v4[6] == 26 * v51 + 67 * v49 + 6 * v47 + 4 * v46 + 3 * v48 + 68 * v52)
s.add(v4[7] == 34 * v56 + 12 * v53 + 53 * v54 + 6 * v55 + 58 * v57 + 36 * v58 + v59)
s.add(v4[8] == 27 * v57 + 73 * v56 + 12 * v55 + 83 * v53 + 85 * v54 + 96 * v58 + 52 * v59)
s.add(v4[9] == 24 * v55 + 78 * v53 + 53 * v54 + 36 * v56 + 86 * v57 + 25 * v58 + 46 * v59)
s.add(v4[10] == 78 * v54 + 39 * v53 + 52 * v55 + 9 * v56 + 62 * v57 + 37 * v58 + 84 * v59)
s.add(v4[11] == 48 * v57 + 14 * v55 + 23 * v53 + 6 * v54 + 74 * v56 + 12 * v58 + 83 * v59)
s.add(v4[12] == 15 * v58 + 48 * v57 + 92 * v55 + 85 * v54 + 27 * v53 + 42 * v56 + 72 * v59)
s.add(v4[13] == 26 * v58 + 67 * v56 + 6 * v54 + 4 * v53 + 3 * v55 + 68 * v59)
s.add(v4[14] == 34 * v63 + 12 * v60 + 53 * v61 + 6 * v62 + 58 * v64 + 36 * v65 + v66)
s.add(v4[15] == 27 * v64 + 73 * v63 + 12 * v62 + 83 * v60 + 85 * v61 + 96 * v65 + 52 * v66)
s.add(v4[16] == 24 * v62 + 78 * v60 + 53 * v61 + 36 * v63 + 86 * v64 + 25 * v65 + 46 * v66)
s.add(v4[17] == 78 * v61 + 39 * v60 + 52 * v62 + 9 * v63 + 62 * v64 + 37 * v65 + 84 * v66)
s.add(v4[18] == 48 * v64 + 14 * v62 + 23 * v60 + 6 * v61 + 74 * v63 + 12 * v65 + 83 * v66)
s.add(v4[19] == 15 * v65 + 48 * v64 + 92 * v62 + 85 * v61 + 27 * v60 + 42 * v63 + 72 * v66)
s.add(v4[20] == 26 * v65 + 67 * v63 + 6 * v61 + 4 * v60 + 3 * v62 + 68 * v66)
s.add(v4[21] == 34 * v70 + 12 * v67 + 53 * v68 + 6 * v69 + 58 * v71 + 36 * v72 + v73)
s.add(v4[22] == 27 * v71 + 73 * v70 + 12 * v69 + 83 * v67 + 85 * v68 + 96 * v72 + 52 * v73)
s.add(v4[23] == 24 * v69 + 78 * v67 + 53 * v68 + 36 * v70 + 86 * v71 + 25 * v72 + 46 * v73)
s.add(v4[24] == 78 * v68 + 39 * v67 + 52 * v69 + 9 * v70 + 62 * v71 + 37 * v72 + 84 * v73)
s.add(v4[25] == 48 * v71 + 14 * v69 + 23 * v67 + 6 * v68 + 74 * v70 + 12 * v72 + 83 * v73)
s.add(v4[26] == 15 * v72 + 48 * v71 + 92 * v69 + 85 * v68 + 27 * v67 + 42 * v70 + 72 * v73)
s.add(v4[27] == 26 * v72 + 67 * v70 + 6 * v68 + 4 * v67 + 3 * v69 + 68 * v73)
s.add(v4[28] == 34 * v77 + 12 * v74 + 53 * v75 + 6 * v76 + 58 * v78 + 36 * v79 + v80)
s.add(v4[29] == 27 * v78 + 73 * v77 + 12 * v76 + 83 * v74 + 85 * v75 + 96 * v79 + 52 * v80)
s.add(v4[30] == 24 * v76 + 78 * v74 + 53 * v75 + 36 * v77 + 86 * v78 + 25 * v79 + 46 * v80)
s.add(v4[31] == 78 * v75 + 39 * v74 + 52 * v76 + 9 * v77 + 62 * v78 + 37 * v79 + 84 * v80)
s.add(v4[32] == 48 * v78 + 14 * v76 + 23 * v74 + 6 * v75 + 74 * v77 + 12 * v79 + 83 * v80)
s.add(v4[33] == 15 * v79 + 48 * v78 + 92 * v76 + 85 * v75 + 27 * v74 + 42 * v77 + 72 * v80)
s.add(v4[34] == 26 * v79 + 67 * v77 + 6 * v75 + 4 * v74 + 3 * v76 + 68 * v80)
s.add(v4[35] == 34 * v84 + 12 * v81 + 53 * v82 + 6 * v83 + 58 * v85 + 36 * v86 + v87)
s.add(v4[36] == 27 * v85 + 73 * v84 + 12 * v83 + 83 * v81 + 85 * v82 + 96 * v86 + 52 * v87)
s.add(v4[37] == 24 * v83 + 78 * v81 + 53 * v82 + 36 * v84 + 86 * v85 + 25 * v86 + 46 * v87)
s.add(v4[38] == 78 * v82 + 39 * v81 + 52 * v83 + 9 * v84 + 62 * v85 + 37 * v86 + 84 * v87)
s.add(v4[39] == 48 * v85 + 14 * v83 + 23 * v81 + 6 * v82 + 74 * v84 + 12 * v86 + 83 * v87)
s.add(v4[40] == 15 * v86 + 48 * v85 + 92 * v83 + 85 * v82 + 27 * v81 + 42 * v84 + 72 * v87)
s.add(v4[41] == 26 * v86 + 67 * v84 + 6 * v82 + 4 * v81 + 3 * v83 + 68 * v87)
print(s.check())
print(s.model())

得到数据经过排序和16进制转化得到flag！

0x04 电脑被黑

Linux命令行调试得到一个flag.txt和一个demo文件

拖入IDA中查看，反编译文件，看逻辑是一个简单的加密。

image-20200821121818907

加密代码就简单几行，分析得到flag.txt为密文

编写简单脚本就得到flag

file=open('flag.txt','rb') 
f=file.read() 
v4=34 
v5=0 
flag="" 
for i in f: 
    flag=flag+chr((ord(i)^v4)-v5) 
    v4=(v4+34)&0xff 
    v5=(v5+2)&0xf 
    #print flag,v4,v5 
print flag

flag{e5d7c4ed-b8f6-4417-8317-b809fc26c047}

0x05 bd

一道RSA题，题目提示是Wiener‘s attack 直接上脚本

import gmpy2
def transform(x,y):       #使用辗转相处将分数 x/y 转为连分数的形式
    res=[]
    while y:
        res.append(x//y)
        x,y=y,x%y
    return res
    
def continued_fraction(sub_res):
    numerator,denominator=1,0
    for i in sub_res[::-1]:      #从sublist的后面往前循环
        denominator,numerator=numerator,i*numerator+denominator
    return denominator,numerator   #得到渐进分数的分母和分子，并返回

    
#求解每个渐进分数
def sub_fraction(x,y):
    res=transform(x,y)
    res=list(map(continued_fraction,(res[0:i] for i in range(1,len(res)))))  #将连分数的结果逐一截取以求渐进分数
    return res

def get_pq(a,b,c):      #由p+q和pq的值通过维达定理来求解p和q
    par=gmpy2.isqrt(b*b-4*a*c)   #由上述可得，开根号一定是整数，因为有解
    x1,x2=(-b+par)//(2*a),(-b-par)//(2*a)
    return x1,x2

def wienerAttack(e,n):
    for (d,k) in sub_fraction(e,n):  #用一个for循环来注意试探e/n的连续函数的渐进分数，直到找到一个满足条件的渐进分数
        if k==0:                     #可能会出现连分数的第一个为0的情况，排除
            continue
        if (e*d-1)%k!=0:             #ed=1 (mod φ(n)) 因此如果找到了d的话，(ed-1)会整除φ(n),也就是存在k使得(e*d-1)//k=φ(n)
            continue
        
        phi=(e*d-1)//k               #这个结果就是 φ(n)
        px,qy=get_pq(1,n-phi+1,n)
        if px*qy==n:
            p,q=abs(int(px)),abs(int(qy))     #可能会得到两个负数，负负得正未尝不会出现
            d=gmpy2.invert(e,(p-1)*(q-1))     #求ed=1 (mod  φ(n))的结果，也就是e关于 φ(n)的乘法逆元d
            return d
    print("该方法不适用")
    
    
e =46867417013414476511855705167486515292101865210840925173161828985833867821644239088991107524584028941183216735115986313719966458608881689802377181633111389920813814350964315420422257050287517851213109465823444767895817372377616723406116946259672358254060231210263961445286931270444042869857616609048537240249
n =86966590627372918010571457840724456774194080910694231109811773050866217415975647358784246153710824794652840306389428729923771431340699346354646708396564203957270393882105042714920060055401541794748437242707186192941546185666953574082803056612193004258064074902605834799171191314001030749992715155125694272289
d=wienerAttack(e,n)
print("d=",d)

算出d= 1485313191830359055093545745451584299495272920840463008756233

然后编写第二个脚本求m

import gmpy2
d= 1485313191830359055093545745451584299495272920840463008756233
c = 37625098109081701774571613785279343908814425141123915351527903477451570893536663171806089364574293449414561630485312247061686191366669404389142347972565020570877175992098033759403318443705791866939363061966538210758611679849037990315161035649389943256526167843576617469134413191950908582922902210791377220066
n = 86966590627372918010571457840724456774194080910694231109811773050866217415975647358784246153710824794652840306389428729923771431340699346354646708396564203957270393882105042714920060055401541794748437242707186192941546185666953574082803056612193004258064074902605834799171191314001030749992715155125694272289
m = pow(c,d,n)
print(hex(m))

m=0x666c61677b64333735323533382d393064302d633337332d636665662d3932343764336531363834387d

经过16进制转化 flag{d3752538-90d0-c373-cfef-9247d3e16848}