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[原创]CTF新手合集：Flag算法分析
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发表于: 2023-12-14 23:12
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为什么是新手合集呢？主要是以下的CTF项目难度都不大，仅限于java层。接下来，我们将详细分析和算法还原下。
这题主要在于算法的还原，代码逻辑是比较清晰的，MainActivity.this.check()为真，.check()函数内字符串满足条件则完成flag.
接下来对算法进行还原：
从代码分析来看，只要循环内满足 ：this.s[i] = (chars[i] ^ 23) 我们用python 来还原这个逻辑：
flag{It_1S_@N_3asY_@nDr0)I)1|d}
这一题的利用到数学里面的二元一次方程来求解。
满足 a.a(editText.getText().toString()) 为真则成功，那么a()函数里面满足这个条件则成功。a[i2] != (b[i2] * iArr[i2] * iArr[i2]) + (c[i2] * iArr[i2]) + d[i2] || a[i2 + 1] != (b[i2] * iArr[i2 + 1] * iArr[i2 + 1]) + (c[i2] * iArr[i2 + 1]) + d[i2]。这个函数 我拆开来开 就是满足：
a[i2]-d[i2] = (b[i2] * iArr[i2] * iArr[i2]) + (c[i2] * iArr[i2])
类似： y = x^2+x，就是一个【一元二次方程】 已知 y 求 x ,再把 x的值拼接起来，就是我们输入的字符串。
算法还原：
flag{MAth_i&_GOOd_DON7_90V_7hInK?
这题对于之前2个来说稍微复杂点。核心思想就是字符MD5加密与给定的字符的比较。不过它难就难在你知道MD5加密的结果，反向推算字符串，但是MD5又是不可逆的，怎么反推回去呢？挺有意思是不，接下来我们分析看看。
要想成功，必需满足红框内的条件为真。接着我们到a()方法中：
输入的字符长度"flag{this_is_a_fake_flag_ahhhhh}"长度相同，满足e().a的条件：
满足 b.a(a(str, str2).toString().substring(0, 4), "utf8").equals(this.a) 以此类推。求解str的值，根据代码逻辑，a(str, str2)字符换算转换后调用b.a（）完成MD5加密返回值 等于 a、b、c等。
通过代码分析，我们还得验证我们的分析结果，万一里面的MD5加密是魔改的，搞了半天不就是很尴尬。
通过hook 验证了我们的分析。通过 a(String str, String str2) 的到的字符串，在切成8段，每段四个字符 MD5 得到的结果就是 a、b、c、d、e、f、g、h字符串。
既然如此 我们就暴力破解，组成相应的字符，进行加密 一一比对就好了。
得到结果：bd1d6ba7f1d3f5a13ebb0a75844cccfa
s = [113, 123, 118, 112, 108, 94, 99, 72, 38, 68, 72, 87, 89, 72, 36, 118, 100, 78, 72, 87, 121, 83, 101, 39, 62, 94, 62, 38, 107, 115, 106]
ss= []
for one in s:
b = one ^ 23
ss.append(b)
ss_b = bytes(ss)
print(ss_b.decode('utf-8')) # flag{It_1S_@N_3asY_@nDr0)I)1|d}
s = [113, 123, 118, 112, 108, 94, 99, 72, 38, 68, 72, 87, 89, 72, 36, 118, 100, 78, 72, 87, 121, 83, 101, 39, 62, 94, 62, 38, 107, 115, 106]
ss= []
for one in s:
b = one ^ 23
ss.append(b)
ss_b = bytes(ss)
print(ss_b.decode('utf-8')) # flag{It_1S_@N_3asY_@nDr0)I)1|d}
from sympy import symbols, Eq, solve
a = [0, 146527998, 205327308, 94243885, 138810487, 408218567, 77866117, 71548549, 563255818, 559010506, 449018203,576200653, 307283021, 467607947, 314806739, 341420795, 341420795, 469998524, 417733494, 342206934, 392460324,382290309, 185532945, 364788505, 210058699, 198137551, 360748557, 440064477, 319861317, 676258995, 389214123,829768461, 534844356, 427514172, 864054312]
b = [13710, 46393, 49151, 36900, 59564, 35883, 3517, 52957, 1509, 61207, 63274, 27694, 20932, 37997, 22069, 8438, 33995,53298, 16908, 30902, 64602, 64028, 29629, 26537, 12026, 31610, 48639, 19968, 45654, 51972, 64956, 45293, 64752,37108]
c = [38129, 57355, 22538, 47767, 8940, 4975, 27050, 56102, 21796, 41174, 63445, 53454, 28762, 59215, 16407, 64340,37644, 59896, 41276, 25896, 27501, 38944, 37039, 38213, 61842, 43497, 9221, 9879, 14436, 60468, 19926, 47198, 8406,64666]
d = [0, -341994984, -370404060, -257581614, -494024809, -135267265, 54930974, -155841406, 540422378, -107286502,-128056922, 265261633, 275964257, 119059597, 202392013, 283676377, 126284124, -68971076, 261217574, 197555158,-12893337, -10293675, 93868075, 121661845, 167461231, 123220255, 221507, 258914772, 180963987, 107841171, 41609001,276531381, 169983906, 276158562]
def solve_quadratic_equation(a, b, c):
# 定义符号变量
x = symbols('x')
# 定义方程
equation = Eq(a * x ** 2 + b * x + c, 0)
# 解方程
solutions = solve(equation, x)
return solutions
print("len(a) ", len(a), "len(c) ", len(c))
my_string = "adksisjhewadksisjhewadksisjhewwwert"
ascii_values = [ord(char) for char in my_string]
iArr = ascii_values
print(iArr)
i = 0
iArr_new = []
while i < len(c):
a1 = a[i]
a2 = a[i + 1]
print("a1 :", a1, " a2 :", a2)
# a1 = (b[i] * iArr[i] * iArr[i]) + (c[i] * iArr[i]) + d[i]
# a2 = (b[i] * iArr[i + 1] * iArr[i + 1]) + (c[i] * iArr[i + 1]) + d[i]
solutions = solve_quadratic_equation(b[i], c[i], d[i] - a1)
sign_v = solutions
print("sign_v : ", sign_v)
iArr_new.append(sign_v)
i = i + 1
iArr_new_str = bytes(iArr_new)
print("iArr_new_str : ", iArr_new_str)
结果 ：[0, 102, 108, 97, 103, 123, 77, 65, 116, 104, 95, 105, 38, 95, 71, 79, 79, 100, 95, 68, 79, 78, 55, 95, 57, 48, 86, 95, 55, 104, 73, 110, 75, 63]
iArr_new_str : b'\x00flag{MAth_i&_GOOd_DON7_90V_7hInK?'
from sympy import symbols, Eq, solve
a = [0, 146527998, 205327308, 94243885, 138810487, 408218567, 77866117, 71548549, 563255818, 559010506, 449018203,576200653, 307283021, 467607947, 314806739, 341420795, 341420795, 469998524, 417733494, 342206934, 392460324,382290309, 185532945, 364788505, 210058699, 198137551, 360748557, 440064477, 319861317, 676258995, 389214123,829768461, 534844356, 427514172, 864054312]
b = [13710, 46393, 49151, 36900, 59564, 35883, 3517, 52957, 1509, 61207, 63274, 27694, 20932, 37997, 22069, 8438, 33995,53298, 16908, 30902, 64602, 64028, 29629, 26537, 12026, 31610, 48639, 19968, 45654, 51972, 64956, 45293, 64752,37108]
c = [38129, 57355, 22538, 47767, 8940, 4975, 27050, 56102, 21796, 41174, 63445, 53454, 28762, 59215, 16407, 64340,37644, 59896, 41276, 25896, 27501, 38944, 37039, 38213, 61842, 43497, 9221, 9879, 14436, 60468, 19926, 47198, 8406,64666]
d = [0, -341994984, -370404060, -257581614, -494024809, -135267265, 54930974, -155841406, 540422378, -107286502,-128056922, 265261633, 275964257, 119059597, 202392013, 283676377, 126284124, -68971076, 261217574, 197555158,-12893337, -10293675, 93868075, 121661845, 167461231, 123220255, 221507, 258914772, 180963987, 107841171, 41609001,276531381, 169983906, 276158562]
def solve_quadratic_equation(a, b, c):
# 定义符号变量
x = symbols('x')
# 定义方程
equation = Eq(a * x ** 2 + b * x + c, 0)
# 解方程
solutions = solve(equation, x)
return solutions
print("len(a) ", len(a), "len(c) ", len(c))
my_string = "adksisjhewadksisjhewadksisjhewwwert"
ascii_values = [ord(char) for char in my_string]
iArr = ascii_values
print(iArr)
i = 0
iArr_new = []
while i < len(c):
a1 = a[i]
a2 = a[i + 1]
print("a1 :", a1, " a2 :", a2)
# a1 = (b[i] * iArr[i] * iArr[i]) + (c[i] * iArr[i]) + d[i]
# a2 = (b[i] * iArr[i + 1] * iArr[i + 1]) + (c[i] * iArr[i + 1]) + d[i]
solutions = solve_quadratic_equation(b[i], c[i], d[i] - a1)
sign_v = solutions
print("sign_v : ", sign_v)
iArr_new.append(sign_v)
i = i + 1
iArr_new_str = bytes(iArr_new)
print("iArr_new_str : ", iArr_new_str)
结果 ：[0, 102, 108, 97, 103, 123, 77, 65, 116, 104, 95, 105, 38, 95, 71, 79, 79, 100, 95, 68, 79, 78, 55, 95, 57, 48, 86, 95, 55, 104, 73, 110, 75, 63]
iArr_new_str : b'\x00flag{MAth_i&_GOOd_DON7_90V_7hInK?'
function hook_so() { Java.perform(function () {
var e = Java.use("com.he.tian.easymix.e");
e["a"].overload('java.lang.String', 'java.lang.String').implementation = function (str, str2) {
console.log('a is called' + ', ' + 'str: ' + str + ', ' + 'str2: ' + str2);
var ret = this.a(str, str2);
console.log('a ret value is ', ret);
return ret;
};
var b = Java.use("com.he.tian.easymix.b");
b["a"].overload('java.lang.String', 'java.lang.String').implementation = function (str, str2) {
console.log('a is called' + ', ' + 'str: ' + str + ', ' + 'str2: ' + str2);
var ret = this.a(str, str2);
console.log('a ret value is ' + ret);
return ret;
};
})
}function hook_so() { Java.perform(function () {
var e = Java.use("com.he.tian.easymix.e");
e["a"].overload('java.lang.String', 'java.lang.String').implementation = function (str, str2) {
console.log('a is called' + ', ' + 'str: ' + str + ', ' + 'str2: ' + str2);
var ret = this.a(str, str2);
console.log('a ret value is ', ret);
return ret;
};
[招生]科锐逆向工程师培训(2024年11月15日实地，远程教学同时开班, 第51期)
最后于 2023-12-14 23:27
被西贝巴巴编辑
，原因： 添加apk 文件
