2019年7月第一周打卡_简单校验writeup

roger

因为是.apk文件，我们直接放入虚拟机里面运行一下，非常简单，也是一个判断注册码的题目

还是用jeb打开分析，我们先看看MainActivity里面的内容

这里通过a方法来判断正确与否，所以我们主要研究a方法，可以看到我们输入的即为v4，因为v4[0] = 0所以我们第一个循环为(1,34)，可以通过：
a[m]== b[m] * v4[m]* v4[m] + c[m] * v4[m] + d[m]
a[m+ 1] == b[m] * v4[m+ 1] * v4[m+ 1] + c[m] * v4[m+ 1] + d[m]
将v4[m]替换为要爆破的i：
(a[m] == b[m] * i * i + c[m] * f + d[m])||(a[m] == b[m-1] * i * i + c[m-1] * i + d[m-1])


脚本如下:

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]

flag = ""
for m in range (1,34):
        for i in range (0,127):
                                        if((a[m] == b[m] * i * i + c[m] * i + d[m])&(a[m] == b[m-1] * i * i + c[m-1] * i + d[m-1])):
                                                flag+=chr(i)
                                            
print(flag)