zoukankan      html  css  js  c++  java
  • [网鼎杯 2020 青龙组]singal

    vm,没有栈操作,找到opcode

    0A 00 00 00 04 00 00 00  10 00 00 00 08 00 00 00
    03 00 00 00 05 00 00 00  01 00 00 00 04 00 00 00
    20 00 00 00 08 00 00 00  05 00 00 00 03 00 00 00
    01 00 00 00 03 00 00 00  02 00 00 00 08 00 00 00
    0B 00 00 00 01 00 00 00  0C 00 00 00 08 00 00 00
    04 00 00 00 04 00 00 00  01 00 00 00 05 00 00 00
    03 00 00 00 08 00 00 00  03 00 00 00 21 00 00 00
    01 00 00 00 0B 00 00 00  08 00 00 00 0B 00 00 00
    01 00 00 00 04 00 00 00  09 00 00 00 08 00 00 00
    03 00 00 00 20 00 00 00  01 00 00 00 02 00 00 00
    51 00 00 00 08 00 00 00  04 00 00 00 24 00 00 00
    01 00 00 00 0C 00 00 00  08 00 00 00 0B 00 00 00
    01 00 00 00 05 00 00 00  02 00 00 00 08 00 00 00
    02 00 00 00 25 00 00 00  01 00 00 00 02 00 00 00
    36 00 00 00 08 00 00 00  04 00 00 00 41 00 00 00
    01 00 00 00 02 00 00 00  20 00 00 00 08 00 00 00
    05 00 00 00 01 00 00 00  01 00 00 00 05 00 00 00
    03 00 00 00 08 00 00 00  02 00 00 00 25 00 00 00
    01 00 00 00 04 00 00 00  09 00 00 00 08 00 00 00
    03 00 00 00 20 00 00 00  01 00 00 00 02 00 00 00
    41 00 00 00 08 00 00 00  0C 00 00 00 01 00 00 00
    07 00 00 00 22 00 00 00  07 00 00 00 3F 00 00 00
    07 00 00 00 34 00 00 00  07 00 00 00 32 00 00 00
    07 00 00 00 72 00 00 00  07 00 00 00 33 00 00 00
    07 00 00 00 18 00 00 00  07 00 00 00 A7 FF FF FF
    07 00 00 00 31 00 00 00  07 00 00 00 F1 FF FF FF
    07 00 00 00 28 00 00 00  07 00 00 00 84 FF FF FF
    07 00 00 00 C1 FF FF FF  07 00 00 00 1E 00 00 00
    07 00 00 00 7A 00 00 00 
    View Code
    with open("a.txt") as f:
        f=f.read().split()
    a=[]
    for i in range(0,len(f),4):
        a.append(eval("0x"+f[i]))
    #print(a)
    v10=0
    v7=0
    v9=0
    v6=0
    while(v10<len(a)):
        if a[v10] == 1:
            print("1:      v4["+str(v7)+"] = v5;")
            v10+=1
            v7+=1
            v9+=1
            continue
        if a[v10] == 2:
            print("2:      v5 = a1["+str(v10 + 1)+"] + v3["+str(v9)+"];")
            v10 += 2
            continue
        if a[v10] == 3:
            print("3:      v5 = v3["+str(v9)+"] - a1["+str(v10 + 1)+"];")
            v10 += 2
            continue
        if a[v10] == 4:
            print("4:      v5 = a1["+str(v10 + 1)+"] ^ v3["+str(v9)+"];")
            v10 += 2
            continue
        if a[v10] == 5:
            print("5:      v5 = a1["+str(v10 + 1)+"] * v3["+str(v9)+"];")
            v10 += 2
            continue
        if a[v10] == 6:
            print("6:    ")
            v10+=1
            continue
        if a[v10] == 7:
            #print("if ( v4[v8] != a1[v10 + 1] ){printf("what a shame...");exit(0);}++v8;v10 += 2;")
            continue
        if a[v10] == 8:
            print("8:      v3["+str(v6)+"] = v5;")
            v10+=1
            v6+=1
            continue
        if a[v10] == 10:
            print("10:     read(v3)")
            v10+=1
            continue
        if a[v10] == 11:
            print("11:     v5 = v3["+str(v9)+"] - 1;")
            v10+=1
            continue
        if a[v10] == 12:
            print("12:     v5 = v3["+str(v9)+"] + 1;")
            v10+=1
            continue
        else:
            continue

    得到处理过程

    10:     read(v3)
    4:      v5 = a1[2] ^ v3[0];
    8:      v3[0] = v5;
    3:      v5 = v3[0] - a1[5];
    1:      v4[0] = v5;
    
    4:      v5 = a1[8] ^ v3[1];
    8:      v3[1] = v5;
    5:      v5 = a1[11] * v3[1];
    1:      v4[1] = v5;
    
    3:      v5 = v3[2] - a1[14];
    8:      v3[2] = v5;
    11:     v5 = v3[2] - 1;
    1:      v4[2] = v5;
    
    12:     v5 = v3[3] + 1;
    8:      v3[3] = v5;
    4:      v5 = a1[21] ^ v3[3];
    1:      v4[3] = v5;
    
    5:      v5 = a1[24] * v3[4];
    8:      v3[4] = v5;
    3:      v5 = v3[4] - a1[27];
    1:      v4[4] = v5;
    
    11:     v5 = v3[5] - 1;
    8:      v3[5] = v5;
    11:     v5 = v3[5] - 1;
    1:      v4[5] = v5;
    
    4:      v5 = a1[34] ^ v3[6];
    8:      v3[6] = v5;
    3:      v5 = v3[6] - a1[37];
    1:      v4[6] = v5;
    
    2:      v5 = a1[40] + v3[7];
    8:      v3[7] = v5;
    4:      v5 = a1[43] ^ v3[7];
    1:      v4[7] = v5;
    
    12:     v5 = v3[8] + 1;
    8:      v3[8] = v5;
    11:     v5 = v3[8] - 1;
    1:      v4[8] = v5;
    
    5:      v5 = a1[50] * v3[9];
    8:      v3[9] = v5;
    2:      v5 = a1[53] + v3[9];
    1:      v4[9] = v5;
    
    2:      v5 = a1[56] + v3[10];
    8:      v3[10] = v5;
    4:      v5 = a1[59] ^ v3[10];
    1:      v4[10] = v5;
    
    2:      v5 = a1[62] + v3[11];
    8:      v3[11] = v5;
    5:      v5 = a1[65] * v3[11];
    1:      v4[11] = v5;
    
    5:      v5 = a1[68] * v3[12];
    8:      v3[12] = v5;
    2:      v5 = a1[71] + v3[12];
    1:      v4[12] = v5;
    
    4:      v5 = a1[74] ^ v3[13];
    8:      v3[13] = v5;
    3:      v5 = v3[13] - a1[77];
    1:      v4[13] = v5;
    
    2:      v5 = a1[80] + v3[14];
    8:      v3[14] = v5;
    12:     v5 = v3[14] + 1;
    1:      v4[14] = v5;
    View Code

    其实v4的值就是opcode中7后面的值,但我是动调出来的

    "22 3f 34 32 72 33 18 a7 31 f1 28 84 c1 1e 7a"
    v4="22 3f 34 32 72 33 18 a7 31 f1 28 84 c1 1e 7a"
    v4=v4.split()
    a1=[10, 4, 16, 8, 3, 5, 1, 4, 32, 8, 5, 3, 1, 3, 2, 8, 11, 1, 12, 8, 4, 4, 1, 5, 3, 8, 3, 33, 1, 11, 8, 11, 1, 4, 9, 8, 3, 32, 1, 2, 81, 8, 4, 36, 1, 12, 8, 11, 1, 5, 2, 8, 2, 37, 1, 2, 54, 8, 4, 65, 1, 2, 32, 8, 5, 1, 1, 5, 3, 8, 2, 37, 1, 4, 9, 8, 3, 32, 1, 2, 65, 8, 12, 1, 7, 34, 7, 63, 7, 52, 7, 50, 7, 114, 7, 51, 7, 24, 7, 167, 7, 49, 7, 241, 7, 40, 7, 132, 7, 193, 7, 30, 7, 122]
    v3=[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
    for i in range(len(v4)):
        v4[i]=eval("0x"+v4[i])
    v3[0]=(v4[0]+a1[5])^a1[2]
    v3[1]=(v4[1]//a1[11])^a1[8]
    v3[2]=v4[2]+1+a1[14]
    v3[3]=(v4[3]^a1[21])-1
    v3[4]=(v4[4]+a1[27])//a1[24]
    v3[5]=v4[5]+2
    v3[6]=(v4[6]+a1[37])^a1[34]
    v3[7]=(v4[7]^a1[43])-a1[40]
    v3[8]=v4[8]
    v3[9]=(v4[9]-a1[53])//a1[50]
    v3[10]=(v4[10]^a1[59])-a1[56]
    v3[11]=v4[11]//a1[65]-a1[62]
    v3[12]=(v4[12]-a1[71])//a1[68]
    v3[13]=(v4[13]+a1[77])^a1[74]
    v3[14]=v4[14]-1-a1[80]
    
    for i in v3:
        print(chr(i),end="")

    得到flag

    >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

    做法二:

    最近在学angr

    import angr
    
    def main():
            p=angr.Project("signal.exe",auto_load_libs=False)
            sm=p.factory.simulation_manager(p.factory.entry_state())
            sm.explore(find=0x40179e,avoid=0x4016e6)
            return sm.found[0].posix.dumps(0)
    if __name__=='__main__':
            print(main())
  • 相关阅读:
    SDUT 2143 图结构练习——最短路径 SPFA模板,方便以后用。。 Anti
    SDUT ACM 1002 Biorhythms 中国剩余定理 Anti
    nyist OJ 119 士兵杀敌(三) RMQ问题 Anti
    SDUT ACM 2157 Greatest Number Anti
    SDUT ACM 2622 最短路径 二维SPFA启蒙题。。 Anti
    二叉索引树 区间信息的维护与查询 Anti
    SDUT ACM 2600 子节点计数 Anti
    UVA 1428 Ping pong 二叉索引树标准用法 Anti
    2010圣诞Google首页效果
    Object
  • 原文地址:https://www.cnblogs.com/harmonica11/p/12987606.html
Copyright © 2011-2022 走看看