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forgot 26 2010-8-14 19:30
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forgot 26 2010-8-14 19:09
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forgot 26 2010-8-14 18:58
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一个干掉杀软的木马。
装了个mse也只能识别为Unruy,没有恢复能力
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forgot 26 2010-8-14 18:23
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一个干掉杀软的木马。
这个是被感染的文件,本体在%windir%\csrss.exe,udp连到nettheif.net注册的一个伪dns网站上。
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forgot 26 2010-8-14 18:14
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一个干掉杀软的木马。
网络神偷+exebinder
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forgot 26 2010-8-12 12:14
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[求助]一个Acm题
邪恶,三进宫
class Node:
    def __init__(self, left=None, right=None, value=None):
        self.left = left
        self.right = right
        self.value = value

def build_child(node, value, path):
    branch, path = path[:1], path[1:]
    if branch == 'L':
        if node.left is None:
            node.left = Node()
        return build_child(node.left, value, path)
    elif branch == 'R':
        if node.right is None:
            node.right = Node()
        return build_child(node.right, value, path)
    else:
        if node.value is None:
            node.value = value
            return True
        else:
            return False

def build_tree(input):
    tree = Node()
    for value, path in input:
        if not build_child(tree, value, path):
            return
    return tree

def is_complete(node):
    if node is None:
        return True
    elif node.value is not None and is_complete(node.left) and is_complete(node.right):
        return True
    else:
        return False

def breadth_first_search(node):
    q = [node]
    path = []
    while q:
        node = q.pop(0)
        if node is None or node in path:
            continue
        path.append(node.value)
        q.append(node.left)
        q.append(node.right)
    return path

def solve(input):
    tree = build_tree(input)
    if tree and is_complete(tree):
        print breadth_first_search(tree)
    else:
        print 'not complete'

def test():
    import re
    s = """    
    (11,LL) (7,LLL) (8,R)
    (5,) (4,L) (13,RL) (2,LLR) (1,RRR) (4,RR) ()
    (3,L) (4,R) ()
    (11,LL) (7,LLL) (8,R) (8,LL)
    (5,) (4,L) (13,RL) (2,LLR) (1,RRR) (4,RR) ()    
    """
    t = []
    for x in re.findall('\((.*?)\)', s):
        if not x:
            solve(t)
            t = []
        else:
            t.append(x.split(','))
            
if __name__ == '__main__':            
    test()
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forgot 26 2010-8-11 09:22
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[求助]反汇编时如何区分数据区和程序区
冯诺依曼机器上这个是不可判定的
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forgot 26 2010-8-10 22:52
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[求助]一个Acm题
没注意看,改一下吧
import re
def solve(input):
    complete = [True]
    #
    # build complete binary tree
    #
    tree = [None]*256
    def put(node, index, path):
        if not path:
            if tree[index] is None:
                tree[index] = node
            else:
                complete[0] = False
        elif path[0] == 'L':
            put(node, 2*index+1, path[1:])
        elif path[0] == 'R':
            put(node, 2*index+2, path[1:])
    for node, path in input:
        put(node, 0, path)
    #
    # check completeness -- all nodes except the root have parent
    #
    
    for i in range(1, len(tree)):
        if tree[i] is not None:
            if tree[(i-1)/2] is None:
                complete[0] = False
                break
    if complete[0]:
        print [x for x in tree if x is not None]
    else:
        print 'not complete'
def solve_all():
    s = """    
    (11,LL) (7,LLL) (8,R)
    (5,) (4,L) (13,RL) (2,LLR) (1,RRR) (4,RR) ()
    (3,L) (4,R) ()
    (11,LL) (7,LLL) (8,R) (8,LL)
    (5,) (4,L) (13,RL) (2,LLR) (1,RRR) (4,RR) ()    
    """
    t = []
    for x in re.findall('\((.*?)\)', s):
        if not x:
            solve(t)
            t = []
        else:
            t.append(x.split(','))
solve_all()
    
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forgot 26 2010-8-10 16:15
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[求助]一个Acm题
作业应该自己做
import re
def solve(input):
    #
    # build complete binary tree
    #
    tree = [None]*256
    def put(node, index, path):
        if not path:
            tree[index] = node
        elif path[0] == 'L':
            put(node, 2*index+1, path[1:])
        elif path[0] == 'R':
            put(node, 2*index+2, path[1:])
    input = [x.split(',') for x in re.findall('\((.*?)\)', input)]
    for node, path in input:
        put(node, 0, path)
    #
    # check completeness -- all nodes except the root have parent
    #
    complete = True
    for i in range(1, len(tree)):
        if tree[i] is not None:
            if tree[(i-1)/2] is None:
                complete = False
                break
    if complete:
        print [x for x in tree if x is not None]
    else:
        print 'not complete'
solve('(11,LL) (7,LLL) (8,R) (5,) (4,L) (13,RL) (2,LLR) (1,RRR) (4,RR)')
solve('(3,L) (4,R)')
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forgot 26 2010-8-8 08:29
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[求助]能帮我解释下吗
得把 string <Module>::a(string, int32)贴出来
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forgot 26 2010-8-7 20:49
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[游戏]猛男测试,挑战 fsg 2.0 脱壳
准备下一轮猛男测试,vmp pure M.
雪    币: 6075
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forgot 26 2010-8-7 17:55
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[讨论]最近滴水调试器,又开始宣传了
那个瓶子可能就是人工替代品
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forgot 26 2010-8-6 23:24
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[游戏]猛男测试,挑战 fsg 2.0 脱壳
贴个好一点的结果
558bec56575368000010006821af4000e8e0f7ffff2bc0a35daf4000a371af4000bea1ae4000ff75
08e821f7ffff400f841108000085c90f84d407000048894608894e0c5051e839f7ffff85c00f84b7
07000085c90f84af0700008906894e049666813e4d5a0f85970700008b463ca903000000751a3b05
adae4000731203f08935b1ae4000813e504500007502eb0a68e8a14000e97c070000837e28007402
eb0a689ca14000e96a07000066817e044c017502eb0a6875a34000e95607000066f7461600200f85
3007000066837e5c010f841e0700008b86f4000000a9000001007502eb0a6894a24000e926070000
85c00f85ef060000817e08465347217402eb0a6875a24000e90907000083be80000000000f84d406
000083bee8000000007502eb0a68a9a34000e9e70600000fb746148d443018a3b5ae40008338000f
84a20600008138555058300f84960600008b4634a3d5ae40008b463ca3c9ae40008b4638a3cdae40
008b4628a3d1ae40000fb75e06891dc5ae40008b15b5ae40008b0dc5ae4000496bc92801ca8b4214
034210e8e4fdffff8b0dadae40003bc876162bc88bf90305a1ae400051506885af4000e85df6ffff
8b4e50c1e102516849af4000e82cf6ffffff35adae40006835af4000e81cf6ffff8b3866c7074d5a
c7473c0c000000c7470c504500008b550c85d274248d4f02b40a813a7b73707d7502eb098a02423c
00740eeb05b02083c204880141fecc75e183c70c893db9ae4000c7470846534721a1d5ae40008947
34c7473c00020000c7475400020000c7477410000000a1cdae4000894738668b460466894704668b
461683c8016689471666c74706020066c74714e0008b461c89471c8b46208947208b46248947248b
462c89472c668b464066894740668b464266894742668b464466894744668b464666894746668b46
4866894748668b464a6689474a8b464c89474c8b46608947608b46648947648b46688947688b466c
89476c668b461866894718668b465c6689475c0fb747148d443818a3bdae40009383631400836310
00c74324e00000c0a1cdae400089430c8b4650e863fcffff89430803430ce858fcffffa3c1ae4000
894334c7433c00020000c7434ce00000c0c7472854010000832519af4000008b86a000000085c074
1150ffb6a400000068eba34000e832fbffff8b86c000000085c0742bff0519af4000e848fcffff97
6a18576801af4000e8e4f4ffffffb6c00000006a186813a44000e8fdfaffff8b86a800000085c074
1150ffb6ac00000068ffa34000e8e2faffff8b467885c0741e50ff767c6871af4000e891dcffffff
7678ff767c6827a44000e8bdfaffff8b35b1ae40000fb70dc5ae4000890dc5ae40008b8688000000
85c0740cffb68c00000050e8d1f6ffffe833ecffffa31daf40008b35b5ae40008b3d49af4000ff35
c5ae40008b46088b561085c075028bc285d2741b8bd88bc2e84ffbffff508b460ce881fbffff5057
e824f4ffff89d8e82ffbffff03f883c628ff0c2475c6592b3d49af400057ff3549af4000e809faff
ff893d55af40008b35b9ae40008b3d35af400081c7000200008b1dc1ae4000833d5daf400000742c
ff3561af4000ff355daf400057e8c7f3ffffa161af4000e8c4faffff899e8800000089868c000000
03f803d8f7451002000000753f833d71af4000007436891d79af40006871af4000e871e9ffffff35
7daf4000ff3571af400057e879f3ffffa17daf4000e876faffff895e7889467c03f803d8ff3595ae
40006a32e831daffff89d80305d5ae4000a3732d400068c82d4000ff3521af4000ff3555af400057
ff3549af4000e86945000083c414e82dfaffff03f803d8ff3599ae40006a32e8eed9ffff833d19af
40000074276a186801af400057e8fff2ffffb818000000e8fcf9ffff899ec00000008986c4000000
03f803d8899e80000000b881000000e8dcf9ffff8986840000008b15d5ae4000a1cdae400003c2a3
6f2d4000b84800000001d803c2a37f2d4000b82800000001d803c2a38b2d4000b84400000001d803
c2a39e2c4000c705832d400080000000c705872d4000007d0000a11daf400003c2a3772d4000be47
2d4000b88f2d4000bae80100008910badc010000895004bade0100008950088b15d1ae400089500c
8b15d5ae4000011001500401500801500c8d4010ba620000008910ba700000008950040118015804
2d472d400001d8890689461068810000005657e809f2ffff81c78100000089fb2b1d35af4000891d
41af400081eb00020000833d85af4000007448f7451001000000753f89d8b900020000e8e0f8ffff
8bd88b3d35af40008dbc3800020000ff3591af4000ff3585af400057e8b0f1ffff033d91af40002b
3d35af4000893d41af40008b35bdae4000895e3889d8e8a0f8ffff8946308b3db9ae400003463489
47508b3d35af400081c754010000be9c2c400068ab0000005657e862f1ffffbba1ae4000ff3541af
4000ff3535af4000ff33e84af1ffffff33ff7304e885f0ffff2bc05050ff3541af4000ff7308e824
380000ff7308e810380000ff7308e8c03700008b3595ae40008b1d99ae4000bf67124000566a34ff
d7ff35adae40006a3ee8f0d7ffff68f4010000e8ed370000536a34ffd7566a36ffd7ff3541af4000
6a40e8cfd7ffff68f4010000e8cc370000536a36ffd7566a38ffd76b0541af40006429d2f735adae
4000ba640000002bd0526847a14000ff3521af4000e8ef3700005883c408506a42ff3555ae4000e8
d13700006858020000e87f370000536a38ffd76a00eb6468cfa24000eb286882a14000eb216801a3
4000eb1a6838a34000eb1368afa14000eb0c6852a24000eb056835a24000bba1ae4000ff33ff7304
e881efffff2bc05050ff730cff7308e823370000ff7308e80f370000ff7308e8bf360000eb05681d
a24000bb6b2b40006849af4000ffd36835af4000ffd3685daf4000ffd36871af4000ffd36821af40
00ffd3585b5f5ec9c20c00
雪    币: 6075
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forgot 26 2010-8-5 09:32
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雪    币: 6075
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forgot 26 2010-8-4 20:56
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雪    币: 6075
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forgot 26 2010-8-4 15:29
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OD插件编写时遇到困难
6楼-----4楼+5楼
雪    币: 6075
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forgot 26 2010-8-4 14:56
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[游戏]猛男测试,挑战 fsg 2.0 脱壳
土方法是什么,我倒很有兴趣知道
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forgot 26 2010-8-4 14:46
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[游戏]猛男测试,挑战 fsg 2.0 脱壳
参考资料
[1] Kahn, A. B. (1962), "Topological sorting of large networks", Communications of the ACM 5 (11): 558–562, doi:10.1145/368996.369025.
[2] Robert Tarjan: Depth-first search and linear graph algorithms. In: SIAM Journal on Computing. Vol. 1 (1972), No. 2, P. 146-160.
[3] Robertson,Edward L. (1977) Code Generation for Short/Long Address Machines.
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