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发布时间:2018-03-30
简要描述:
本文作者:tinyfisher原创投稿活动:...
本文作者:tinyfisher
还是遍历,只不过flag长度变成了27位,去掉“flag{}”6位,因此中间长度为21位,也就是0-0b111111111111111111111修改一下长度就可以
def lfsr(R,mask): output = (R 1) 1)^out if i==0: if tmp==0xB2: pass else: break if i==1: if tmp==0xE9: pass else: break if i==2: if tmp==0x0E: pass else: break if i==3: if tmp==0x13: pass else: break if i==4: if tmp==0xA0: print tmpr else: break
换汤不换药,虽说是1024X1024,但flag长度还是固定的21位:
def nlfsr(R,mask): output = (R 1) 1)^out if i==0: if tmp==0xD1: pass else: break if i==1: if tmp==0xD9: pass else: break if i==2: if tmp==0x40: pass else: break if i==3: if tmp==0x43: pass else: break if i==4: if tmp==0x93: print tmpr else: break
这题需要让函数a返回true,传递的参数paramString为flag,需要我们逆出flag,算法大概的意思:
if (paramString.length() != b.length) { return false; }
这里说明了flag的长度需要等于b数组的长度,也就是34,再往下看:
int[] arrayOfInt = new int[a.length]; arrayOfInt[0] = 0; byte[] arrayOfByte = paramString.getBytes(); int i = arrayOfByte.length; int j = 0; int k = 1; while (j i) { arrayOfInt[k] = arrayOfByte[j]; k++; j++; }
for (int m = 0;; m++) { if (m >= c.length) { break label175; } if ((a[m] != b[m] * arrayOfInt[m] * arrayOfInt[m] + c[m] * arrayOfInt[m] + d[m]) || (a[(m + 1)] != b[m] * arrayOfInt[(m + 1)] * arrayOfInt[(m + 1)] + c[m] * arrayOfInt[(m + 1)] + d[m])) { break; } }
由于if里面是||,也就是0||0才为0,转换一下这个条件就是:
a[m] == b[m] * arrayOfInt[m] * arrayOfInt[m] + c[m] * arrayOfInt[m] + d[m]
a[(m + 1)] == b[m] * arrayOfInt[(m + 1)] * arrayOfInt[(m + 1)] + c[m] * arrayOfInt[(m + 1)] + d[m]
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 f1 in range(32,127): if((a[m] == b[m-1] * f1 * f1 + c[m-1] * f1 + d[m-1]) and (a[m] == b[m] * f1 * f1 + c[m] * f1 + d[m])): flag+=chr(f1) break else: pass #print len(c) print flag+"}"
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