常见简单加密算法逆向

做题

[HUBUCTF 2022 新生赛]simple_RE

base64换表

没有研究清楚具体过程，技巧是看到$\mathrm{0x30}$、$\mathrm{0x3C}$、$\mathrm{0x3F}$即为base64（可能换表）

使用CyberChef进行解码

[SWPUCTF 2022 新生赛]base64

base64解码

标准表，直接解码

[HDCTF 2023]easy_re

upx脱壳，base64解密

[LitCTF 2023]enbase64

base64换表

换表过程比较复杂，研究算法得表。

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#include <cstdio>
#include <cstring>
using namespace std;
int v3[65];
char Source[65]="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/",Destination[65];
int main(void){
    v3[0] = 16;
	//...
    strcpy(Destination,Source);
    for(register int i=0;i<=47;i++){
        for(register int j=0;j<=63;j++)
            Source[j]=Destination[v3[j]];
        strcpy(Destination,Source);
    };
    for(register int i=0;i<=65;i++)
        printf("%c",Destination[i]);
    return 0;
};

[LitCTF 2023]enbase64

base64换表

换表过程比较复杂，研究算法得表。

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#include <cstdio>
#include <cstring>
using namespace std;
int v3[65];
char Source[65]="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/",Destination[65];
int main(void){
    v3[0] = 16;
	//...
    strcpy(Destination,Source);
    for(register int i=0;i<=47;i++){
        for(register int j=0;j<=63;j++)
            Source[j]=Destination[v3[j]];
        strcpy(Destination,Source);
    };
    for(register int i=0;i<=65;i++)
        printf("%c",Destination[i]);
    return 0;
};

[NSSCTF 2022 Spring Recruit]easy Python

python-base64

没啥难的，见到了base64的python写法，然后重新把base64详细学了一遍

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def encode(string,string2):
    tmp_str = str()
    ret = str()
    bit_string_str = string.encode()
    remain = len( string ) % 3
    remain_str = str()
    for char in bit_string_str:
        b_char = (bin(char)[2:])
        b_char = '0'*(8-len(b_char)) + b_char
        tmp_str += b_char
    for i in range(len(tmp_str)//6):
        temp_nub = int(tmp_str[i*6:6*(i+1)],2)
        ret += string2[temp_nub]
    if remain==2:
        remain_str = tmp_str[-4:] + '0'*2
        temp_nub = int(remain_str,2)
        ret += string2[temp_nub] + "="
    elif remain==1:
        remain_str = tmp_str[-2:] + '0'*4
        temp_nub = int(remain_str,2)
        ret += string2[temp_nub] + "="*2
    return ret.replace("=","")

[LitCTF 2023]ez_XOR

简单异或，就是反编译代码比较销魂

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s='E`}J]OrQF[V8zV:hzpV}fVF[t'
for i in s:
    i=chr(ord(i)^9)
    print(i,end='')

[HNCTF 2022 Week1]贝斯是什么乐器啊？

简单

[羊城杯 2020]easyre

盲猜base64->打乱->自定义加密，exp:

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#include <cstdio>
#include <cstring>
#include <cctype>
using namespace std;
char str[]="EmBmP5Pmn7QcPU4gLYKv5QcMmB3PWHcP5YkPq3=cT6QckkPckoRG",flag[100],flag2[100],flag3[100];
int len;
int main(void){
    len=strlen(str);
    for(register int i=0;i<len;i++){
        if(isdigit(str[i])==true){
            if((str[i]>='0')&&(str[i]<='2'))
                flag[i]=str[i]+7;
            else
                flag[i]=str[i]-3;
            continue;
        };
        if((str[i]>='a')&&(str[i]<='z')){
            if((str[i]>='a')&&(str[i]<='c'))
                flag[i]=str[i]+23;
            else
                flag[i]=str[i]-3;
            continue;
        };
        if((str[i]>='A')&&(str[i]<='Z')){
            if((str[i]>='A')&&(str[i]<='C'))
                flag[i]=str[i]+23;
            else
                flag[i]=str[i]-3;
            continue;
        };
    };
    strncpy(flag2+13,flag+39,13);
    strncpy(flag2+39,flag+26,13);
    strncpy(flag2,flag+13,13);
    strncpy(flag2+26,flag,13);
    printf("%s\n",flag2);
    return 0;
};

然后flag2解密base64

[SWPUCTF 2022 新生赛]base64-2

不知道咋做出来的，找了个像base64的、找了个像表的，就解出来了…

[GWCTF 2019]pyre

exp:

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code = ['\x1f', '\x12', '\x1d', '(', '0', '4', '\x01', '\x06', '\x14', '4', ',', '\x1b', 'U', '?', 'o', '6', '*', ':', '\x01', 'D', ';', '%', '\x13']
l=len(code)
for i in range(l-2,-1,-1):
    code[i]=chr(ord(code[i])^ord(code[i+1]))
for i in range(l):
    if ord(code[i])-i>=1:
        code[i]=chr(ord(code[i])-i)
    else:
        code[i]=chr(ord(code[i])+128-i)
flag="".join(code)
print(flag)

[MoeCTF 2022]chicken_soup

去花指令、创建函数进行反编译，strings找flag

作为切入点的比较函数比较抽象，提取答案数组数据

！！！GET新技能：

该语句可逆：

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>(a<<4)|(a>>4)

还有简单变换，exp:

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#include <cstdio>
using namespace std;
int map[]={0xCD,0x4D,0x8C,0x7D,0xAD,0x1E,0xBE,0x4A,0x8A,0x7D,0xBC,0x7C,0xFC,0x2E,0x2A,0x79,0x9D,0x6A,0x1A,0xCC,0x3D,0x4A,0xF8,0x3C,0x79,0x69,0x39,0xD9,0xDD,0x9D,0xA9,0x69,0x4C,0x8C,0xDD,0x59,0xE9,0xD7},ans1[36];
int main(void){
    for(register int i=0;i<38;i++)
        map[i]=(map[i]<<4)|(map[i]>>4);
    for(register int i=36;i>=0;i--)
        map[i]-=map[i+1];
    for(register int i=0;i<38;i++)
        printf("%c",map[i]);
    return 0;
};

[WUSTCTF 2020]level1

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#include <cstdio>
using namespace std;
int map[20]={0,198,232,816,200,1536,300,6144,984,51200,570,92160,1200,565248,756,1474560,800,6291456,1782,65536000};
int main(void){
    for(register int i=1;i<=19;i++)
        if((i&1)==1)
            printf("%c",map[i]>>i);
        else
            printf("%c",map[i]/i);
    return 0;
};

[HGAME 2023 week2]before_main

sub_12EB()为base64编码，qword_4020为表，交叉引用发现表被sub_1228()篡改，原因是init()中off_3D78有dq offset sub_1228+1

[HGAME 2023 week1]easyenc

难度不大的加密，两个坑点

  • c语言有溢出…chr()之前要&0xFF
  • DIE查壳为LE字节序…每4位反一下

[Week 3] 旋转密码城

有会的了

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#include <cstdio>
#include <cstring>
using namespace std;
char enc[45]="_Iv2>6L424c_4c2\\f__5\\7fec\\da32\\3ef2`cgd4b46N",rec[45];
void dfs(const int k){
    if(k==45){
        for(register int i=0;i<44;i++)
            printf("%c",rec[i]);
        printf("\n");
        return;
    };
    if((enc[k]<'!')||(enc[k]>'~')){
        rec[k]=enc[k];
        dfs(k+1);
    };
    char tmp1=enc[k]-33-14;
    if((tmp1>='!')&&(tmp1<='~')){
        rec[k]=enc[k]-33-14;
        dfs(k+1);
    };
    char tmp2=enc[k]-33-94-14;
    if((tmp2>='!')&&(tmp2<='~')){
        rec[k]=enc[k]-33-94-14;
        dfs(k+1);
    };
    char tmp3=enc[k]-33+94-14;
    if((tmp3>='!')&&(tmp3<='~')){
        rec[k]=enc[k]-33+94-14;
        dfs(k+1);
    };
    return;
};
int main(void){
    dfs(0);
    return 0;
};

[AFCTF 2018]凯撒

分析主要代码逻辑，发现是个不太被发现的凯撒加密…

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#include <cstdio>
using namespace std;
int map[24]={
    0x6C, 0x71, 0x6E, 0x7F, 0x71, 0x86, 0x6E, 0x3C, 0x3F, 0x7E, 0x7E, 0x74, 0x6E, 0x6A, 0x6E, 0x6C, 
    0x70, 0x40, 0x6C, 0x7D, 0x88, 0x00, 0x00, 0x00
};
int main(void){
    for(register int i=0;i<=16;i++){
        for(register int j=0;j<21;j++)
            printf("%c",map[j]-i);
        putchar('\n');
    };
    return 0;
};

[鹏城杯 2022]baby_re

简单逻辑。

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key=[0x56^0x47,0x57^0x32,0x58^0x11,0x59^0x12]
enc=[119,9,40,44,106,83,126,123,33,87,113,123,112,93,125,127,41,82,44,127,39,3,126,125,119,87,47,125,33,6,44,127,112,0,126,123,115,24]
for i in range(len(enc)):
    enc[i]^=key[i%4]
    print(chr(enc[i]),end='')

[羊城杯 2021]Ez_android

用户名可知为‘admin’，密码每两位转字节减1，查询MD5，得口令654321。连接主机，取得Base64表，解码即得。这是一部分exp：

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enc='3lkHi9iZNK87qw0p6U391t92qlC5rwn5iFqyMFDl1t92qUnL6FQjqln76l-P'
passwd='c232666f1410b3f5010dc51cec341f58'
for i in range(0,len(passwd),2):
    tmp1=passwd[i]+passwd[i+1]
    tmp2=int(tmp1,16)
    tmp2+=1
    print("%02x"%tmp2,end='')
#654321

[MoeCTF 2022]fake_key

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enc=[0x15,0x21,0x0F,0x19,0x25,0x5B,0x19,0x39,0x5F,0x3A,0x3B,0x30,0x74,0x07,0x43,0x3F,0x09,0x5A,0x34,0x0C,0x74,0x3F,0x1E,0x2D,0x27,0x21,0x12,0x16,0x1F]
ans1=[0x18,0x14,0x0F,0x1B,0x09,0x50,0x0B,0x14,0x0F,0x35,0x22,0x2D,0x4D,0x15,0x13,0x00,0x02,0x0A,0x38,0x38,0x25,0x32,0x18,0x14,0x0F,0x1B,0x09,0x50,0x0B]
key=list(b"yunzh1junTCL,trackYYDS")
ans2=[0x19,0x1B,0x13,0x1B,0x12,0x54,0x13,0x1C,0x11,0x39,0x27,0x32,0x4E,0x1C,0x14,0x01,0x07,0x0C,0x3F,0x3E,0x26,0x36,0x1A,0x17,0x11,0x1D,0x0A,0x56,0x13]
rnd=[0]*29
for i in range(len(ans1)):
    rnd[i]=ans2[i]-ans1[i]
for i in range(len(enc)):
    enc[i]-=rnd[i]
for i in range(len(ans1)):
    enc[i]^=key[i%len(key)]
for i in range(len(enc)):
    print(chr(enc[i]),end='')

[WUSTCTF 2020]level4

已知中序及后序求前序遍历。

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def PreOrder(Postorder,Inorder):
    length = len(Postorder)
    if length == 0:
        return 0
    root = Postorder[length-1]
    for i in range(length):
        if root == Inorder[i]:
            break
    print(root,end="")
    PreOrder(Postorder[0:i],Inorder[0:i])
    PreOrder(Postorder[i:length-1],Inorder[i+1:length])
PreOrder("20f0Th{2tsIS_icArE}e7__w","2f0t02T{hcsiI_SwA__r7Ee}")

[MoeCTF 2021]EinfachRe

一开始在疯狂枚举，突然想起来Moe的flag的前7个字符好像已经确定了哈~

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enc=[0x28,0x15,0x3A,0x1B,0x44,0x14,0x06]
known="moectf{"
print(known,end='')
for i in range(len(enc)):
    print(chr(enc[i]^ord(known[i])),end='')

[FSCTF 2023]Xor

大水题

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enc=[0x74,0x7e,0x73,0x75,0x69,0x46,0x7a,0x7b,0x61,0x4d,0x7b,0x61,0x4d,0x73,0x4d,0x77,0x73,0x61,0x6b,0x4d,0x60,0x77,0x64,0x77,0x60,0x61,0x77,0x6f,0x12]
for i in range(len(enc)):
    print(chr(enc[i]^0x12),end='')

[SDCTF 2022]A Bowl of Pythons

简单斐波那契XOR加密，题目很新颖。

题面：

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#! /usr/bin/env python3
FLAG = 'sdctf{a_v3ry_s3cur3_w4y_t0_st0r3_ur_FLAG}' # lol

a = lambda n: a(n-2) + a(n-1) if n >= 2 else (2 if n == 0 else 1)

b = lambda x: bytes.fromhex(x).decode()

h = eval(b('7072696e74'))#print

def d():
    h(b('496e636f727265637420666c61672120596f75206e65656420746f206861636b206465657065722e2e2e')) #Incorrect flag! You need to hack deeper...
    eval(b('5f5f696d706f72745f5f282273797322292e65786974283129')) #__import__("sys").exit(1)
    h(FLAG)

def e(f):
    h("Welcome to SDCTF's the first Reverse Engineering challenge.")
    c = input("Input the correct flag: ")
    if c[:6].encode().hex() != '{2}3{0}{1}{0}3{2}{1}{0}{0}{2}b'.format(*map(str, [6, 4, 7])): #sdctf{
        d()
    if c[int(chr(45) + chr(49))] != chr(125): #}
        d()
    g = c[6:-1].encode()
    if bytes( (g[i] ^ (a(i) & 0xff) for i in range(len(g))) ) != f:
        d()
    h(b('4e696365206a6f622e20596f7520676f742074686520636f727265637420666c616721')) #Nice job. You got the correct flag!

if __name__ == "__main__":
    e(b'co\\7\x7f\x7f`|p\x15\x0e\x8a\x0fP\x14\x18\xfe\xa9\xf3\xe2y\xdd')
else:
    eval(b('5f5f696d706f72745f5f282273797322292e65786974283029'))

exp：

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s=b'co\\7\x7f\x7f`|p\x15\x0e\x8a\x0fP\x14\x18\xfe\xa9\xf3\xe2y\xdd'
s=list(s[i]for i in range(len(s)))
a=lambda n:a(n-2)+a(n-1)if n>=2 else(2 if n==0 else 1)
for i in range(len(s)):
    s[i]^=(a(i)&0xff)
for i in range(len(s)):
    print(chr(s[i]),end='')

[SUCTF 2018 招新赛]hash

看到有MD5常数，猜测为MD5加密，然后还有一些变形。另一半flag在自定义加密中。

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cmp="`ut9t;"
md5="bf772f6ed89838b9gb9f7abf3cc09413"
c_md5=""
c_cmp=""
for i in range(len(md5)):
    if((i&1)==0):
        c_md5+=chr(ord(md5[i])^1)
    else:
        c_md5+=md5[i]
for j in range(len(cmp)):
    c_cmp+=chr(ord(cmp[j])-2*j-1)
print(c_md5)
print(c_cmp)

MD5解码后拼起来即可。

[MoeCTF 2021]ez_Algorithm

程序递归层数太大，考虑递归转递推：

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#include <cstdio>
using namespace std;
const int xorkey[76]={
    0x0000006D, 0x0000006E, 0x00000060, 0x000000C8, 0x00005521, 0x00AAAACD, 0x5555552E, 0xAAAAAA9F, 
    0x55555533, 0xAAAAAA9C, 0x55555566, 0xAAAAAAD9, 0x5555550A, 0xAAAAAACA, 0x55555564, 0xAAAAAA9A, 
    0x5555550A, 0xAAAAAA9C, 0x5555553D, 0xAAAAAA9A, 0x55555526, 0xAAAAAAF4, 0x55555562, 0xAAAAAA9A, 
    0x55555538, 0xAAAAAA98, 0x5555552B, 0xAAAAAAEF, 0x55555565, 0xAAAAAAF4, 0x5555552C, 0xAAAAAA9B, 
    0x55555520, 0xAAAAAAF4, 0x55555527, 0xAAAAAA98, 0x55555534, 0xAAAAAA9A, 0x55555564, 0xAAAAAAD1, 
    0x55555566, 0xAAAAAAF4, 0x55555562, 0xAAAAAAC3, 0x55555566, 0xAAAAAAF4, 0x55555538, 0xAAAAAA98, 
    0x55555534, 0xAAAAAAC5, 0x55555564, 0xAAAAAAC5, 0x55555532, 0xAAAAAAF4, 0x55555565, 0xAAAAAACD, 
    0x5555550A, 0xAAAAAAFF, 0x55555564, 0xAAAAAAC6, 0x55555566, 0xAAAAAAF4, 0x55555536, 0xAAAAAA9B, 
    0x55555538, 0xAAAAAADB, 0x55555564, 0xAAAAAA98, 0x5555552D, 0xAAAAAA9A, 0x55555562, 0xAAAAAAD2, 
    0x5555556A, 0xAAAAAA94, 0x5555556A, 0xAAAAAAD6
};
int dp[5626];
int main(void){
    dp[1]=1;
    for(register int i=2;i<=5625;i++)
        dp[i]=dp[i-1]+2*dp[i-2];
    for(register int i=0;i<=75;i++)
        dp[i*i]^=xorkey[i];
    for(register int i=0;i<=75;i++)
        printf("%c",dp[i*i]);
    return 0;
};

[鹏城杯 2022]rota

加了一大堆优化。算法流程：变表Base64、自定义加密。

自定义加密为单字节加密，直接爆破即可。

自定义加密的密钥需要动调。

脚本是网上wp的，自己写不出来。

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import base64
key=[0x33,0x34,0x2C,0x36,0x1D,0x12,0x1E,0x0C,0x1A,0x3C,0x29,0x10,0x20,0x14,0x3D,0x3B,0x19,0x08,0x0E,0x1F,0x30,0x05,0x38,0x03,0x11,0x1B,0x17,0x21,0x2E,0x04,0x18,0x23,0x2B,0x02,0x27,0x37,0x1C,0x24,0x39,0x3F,0x35,0x2D,0x26,0x13,0x2A,0x0A,0x00,0x07,0x3E,0x01,0x28,0x2F,0x32,0x22,0x0D,0x06,0x25,0x3A,0x09,0x0F,0x16,0x0B,0x15,0x31,0x0C,0x2C,0x0D,0x21,0x22,0x09,0x02,0x39,0x31,0x17,0x1A,0x33,0x06,0x24,0x10,0x04,0x1B,0x0B,0x34,0x12,0x38,0x27,0x0E,0x20,0x2B,0x2E,0x00,0x13,0x3E,0x3A,0x05,0x1E,0x36,0x08,0x32,0x29,0x19,0x23,0x3D,0x3B,0x3C,0x3F,0x37,0x30,0x18,0x16,0x35,0x25,0x0A,0x2D,0x28,0x26,0x15,0x11,0x07,0x1D,0x2A,0x0F,0x1F,0x14,0x01,0x1C,0x03,0x2F,0x13,0x0D,0x35,0x31,0x07,0x11,0x1B,0x23,0x0B,0x0C,0x10,0x25,0x2B,0x21,0x33,0x18,0x27,0x29,0x02,0x2F,0x28,0x30,0x0E,0x19,0x3C,0x08,0x34,0x20,0x3D,0x2E,0x05,0x15,0x2C,0x1C,0x36,0x22,0x1E,0x24,0x38,0x0A,0x3F,0x1A,0x04,0x26,0x16,0x2A,0x3A,0x1F,0x2D,0x32,0x06,0x37,0x03,0x3B,0x00,0x17,0x1D,0x12,0x09,0x01,0x3E,0x39,0x0F,0x14]
compare_data='ksPhS/34MXifj+Ibtjud2Tikj5HkA7iTpbaNELBebOaIm'
base64_table='XiIzDuAoGlaK6JcjM3g/9YQmHBOsxn1hLZ4w7Tt0PV5pNqUFC+rE2dSfyvWe8kRb'
base64_standard_table='ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/'
flag_tmp=''
for i in range(45):
    for str in base64_table:
        index=base64_table.find(str)
        v12=key[(i+index)&0x3f]&0x3f
        v14=key[(key[v12+64]&0x3f)+128]
        if base64_table[v14] == compare_data[i]:
            flag_tmp+=str
            break
flag_tmp=flag_tmp[:44]
print(base64.b64decode(flag_tmp.translate(str.maketrans(base64_table,base64_standard_table))))

[GXYCTF 2019]minecraft

NSS上这题flag怎么交也不对…

.exe中关键函数String_to_long为.dll中的，.dll中逻辑为标准Base64+Hash。其中Hash函数藏得很深，不看wp根本找不到。

exp不是自己写的：

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#include <cstdio>
unsigned int enc[8]={0x6C43B2A7,0x7954FD91,0xA3E9532,0xB87B5156,0xDA847742,0x2395E7F3,0xA679D954,0xE1FAAFF7};
char table[69]="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789{}_-";
bool flag;
int main(void){
    for(int i=0;i<8;i++){
        flag=true;
        for(int n1=0;n1<66&&flag;n1++)
            for(int n2=0;n2<66&&flag;n2++)
                for(int n3=0;n3<66&&flag;n3++)
                    for(int n4=0;n4<66&&flag;n4++){
                        unsigned long long int res=0;
                        res ^= 0xC6A4A7935BD1E995LL * (((unsigned long long)(0xC6A4A7935BD1E995LL * table[n1]) >> 47) ^ 0xC6A4A7935BD1E995LL * table[n1]);
                        res *= 0xC6A4A7935BD1E995LL;
                        res += 0xE6546B64;
                        res ^= 0xC6A4A7935BD1E995LL * (((unsigned long long)(0xC6A4A7935BD1E995LL * table[n2]) >> 47) ^ 0xC6A4A7935BD1E995LL * table[n2]);
                        res *= 0xC6A4A7935BD1E995LL;
                        res += 0xE6546B64;
                        res ^= 0xC6A4A7935BD1E995LL * (((unsigned long long)(0xC6A4A7935BD1E995LL * table[n3]) >> 47) ^ 0xC6A4A7935BD1E995LL * table[n3]);
                        res *= 0xC6A4A7935BD1E995LL;
                        res += 0xE6546B64;
                        res ^= 0xC6A4A7935BD1E995LL * (((unsigned long long)(0xC6A4A7935BD1E995LL * table[n4]) >> 47) ^ 0xC6A4A7935BD1E995LL * table[n4]);
                        res *= 0xC6A4A7935BD1E995LL;
                        res += 0xE6546B64;
                        if((unsigned int)res==enc[i]){//这里比较坑_int64转int
                            flag=false;
                            printf("%c%c%c%c",table[n1],table[n2],table[n3],table[n4]);
                        };
                    };
    };
    return 0;
};

[AFCTF 2018]简单编码

进行一个爆破

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enc=[0x64,0x75,0x6a,0x48,0x4d,0xe7,0x6d,0x2a,0xea,0xf2,0x8f,0x2f,0x6c,0x8a,0xdc]
alb="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789_-"
def enc1(word, i):
        word = ord(word) ^ 0x76 ^ 0xAD
        temp1 = (word & 0xAA) >> 1
        temp2 = 2 * word & 0xAA
        word = temp1 | temp2
        return word
def enc2(word, i):        
        word = ord(word) ^ 0x76 ^ 0xBE
        temp1 = (word & 0xCC) >> 2
        temp2 = 4 * word & 0xCC
        word = temp1 | temp2
        return word
def enc3(word,i):
        word = ord(word) ^ 0x76 ^ 0xEF
        temp1 = (word & 0xF0) >> 4
        temp2 = 16 * word & 0xF0
        word = temp1 | temp2
        return word 
for i in range(5):
    for ch in alb:
        if(enc1(ch,i)==enc[i]):
            print(ch,end='')
            break
for i in range(5):
    for ch in alb:
        if(enc2(ch,i)==enc[i+5]):
            print(ch,end='')
            break
for i in range(5):
    for ch in alb:
        if(enc3(ch,i)==enc[i+10]):
            print(ch,end='')
            break

刮开有奖

密钥爆破代码：

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#include <cstdio>
#include <cstring>
void __cdecl sub_4D10F0(char *a1, int a2, int a3)
{
    int v3; // eax
    int i;  // esi
    int v5; // ecx
    int v6; // edx

    v3 = a3;
    for (i = a2; i <= a3; a2 = i)
    {
        v5 = i;
        v6 = a1[i];
        if (a2 < v3 && i < v3)
        {
            do
            {
                if (v6 > a1[v3])
                {
                    if (i >= v3)
                        break;
                    ++i;
                    a1[v5] = a1[v3];
                    if (i >= v3)
                        break;
                    while (a1[i] <= v6)
                    {
                        if (++i >= v3)
                            goto LABEL_13;
                    }
                    if (i >= v3)
                        break;
                    v5 = i;
                    a1[v3] = a1[i];
                }
                --v3;
            } while (i < v3);
        }
    LABEL_13:
        a1[v3] = v6;
        sub_4D10F0(a1, a2, i - 1);
        v3 = a3;
        ++i;
    }
}
char str[10];
int main(void)
{
    strcpy(str, "ZJSECaNH3ng");
    sub_4D10F0(str, 0, 10);
    for (int i = 0; i < 11; i++)
        printf("%c", str[i]);
}

[羊城杯 2023]Blast

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import hashlib
charset='ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789_!()'
hashset1=[]
hashset2=[]
for i in range(len(charset)):
    m=hashlib.md5()
    m.update(bytes(charset[i].encode()))
    hashset1.append(m.hexdigest())
for i in range(len(hashset1)):
    m=hashlib.md5()
    m.update(bytes(hashset1[i].encode()))
    hashset2.append(m.hexdigest())
cmp=['14d89c38cd0fb23a14be2798d449c182','a94837b18f8f43f29448b40a6e7386ba','af85d512594fc84a5c65ec9970956ea5','af85d512594fc84a5c65ec9970956ea5','10e21da237a4a1491e769df6f4c3b419','a705e8280082f93f07e3486636f3827a','297e7ca127d2eef674c119331fe30dff','b5d2099e49bdb07b8176dff5e23b3c14','83be264eb452fcf0a1c322f2c7cbf987','a94837b18f8f43f29448b40a6e7386ba','71b0438bf46aa26928c7f5a371d619e1','a705e8280082f93f07e3486636f3827a','ac49073a7165f41c57eb2c1806a7092e','a94837b18f8f43f29448b40a6e7386ba','af85d512594fc84a5c65ec9970956ea5','ed108f6919ebadc8e809f8b86ef40b05','10e21da237a4a1491e769df6f4c3b419','3cfd436919bc3107d68b912ee647f341','a705e8280082f93f07e3486636f3827a','65c162f7c43612ba1bdf4d0f2912bbc0','10e21da237a4a1491e769df6f4c3b419','a705e8280082f93f07e3486636f3827a','3cfd436919bc3107d68b912ee647f341','557460d317ae874c924e9be336a83cbe','a705e8280082f93f07e3486636f3827a','9203d8a26e241e63e4b35b3527440998','10e21da237a4a1491e769df6f4c3b419','f91b2663febba8a884487f7de5e1d249','a705e8280082f93f07e3486636f3827a','d7afde3e7059cd0a0fe09eec4b0008cd','488c428cd4a8d916deee7c1613c8b2fd','39abe4bca904bca5a11121955a2996bf','a705e8280082f93f07e3486636f3827a','3cfd436919bc3107d68b912ee647f341','39abe4bca904bca5a11121955a2996bf','4e44f1ac85cd60e3caa56bfd4afb675e','45cf8ddfae1d78741d8f1c622689e4af','3cfd436919bc3107d68b912ee647f341','39abe4bca904bca5a11121955a2996bf','4e44f1ac85cd60e3caa56bfd4afb675e','37327bb06c83cb29cefde1963ea588aa','a705e8280082f93f07e3486636f3827a','23e65a679105b85c5dc7034fded4fb5f','10e21da237a4a1491e769df6f4c3b419','71b0438bf46aa26928c7f5a371d619e1','af85d512594fc84a5c65ec9970956ea5','39abe4bca904bca5a11121955a2996bf']
for i in range(len(cmp)):
    print(charset[hashset2.index(cmp[i])],end='')

[SWPUCTF 2023 秋季新生赛]Char And Int

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enc=[0x43535344,0x667B46EF,0x633635A9,0x2D36BCBC,0x626CDAD6,0x65C8C8D2,0x396BC3DC,0x342DE9E4,0x38396DFA,0x623737DD,0x7D38393E]
keys=[0xa,0xbb,0xccc,0xdddd,0xeeeee,0xffffff,0xeeeee,0xdddd,0xccc,0xbb,0xa]
for i in range(len(enc)):
    enc[i]^=keys[i]
    print(str(hex(enc[i]))[2:],end='')

[CISCN 2022 东北]happymath

直接爆破，脑细胞烧没了。

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#include <cstdio>
using namespace std;
const char c[17]={0x30,0x31,0x32,0x33,0x34,0x35,0x36,0x37,0x38,0x39,0x61,0x62,0x63,0x64,0x65,0x66,0};
unsigned int str1[4096]={0x0,0xf26b8303,0xe13b70f7,0x1350f3f4,0xc79a971f,0x35f1141c,0x26a1e7e8,0xd4ca64eb,0x8ad958cf,0x78b2dbcc,0x6be22838,0x9989ab3b,0x4d43cfd0,0xbf284cd3,0xac78bf27,0x5e133c24,0x105ec76f,0xe235446c,0xf165b798,0x30e349b,0xd7c45070,0x25afd373,0x36ff2087,0xc494a384,0x9a879fa0,0x68ec1ca3,0x7bbcef57,0x89d76c54,0x5d1d08bf,0xaf768bbc,0xbc267848,0x4e4dfb4b,0x20bd8ede,0xd2d60ddd,0xc186fe29,0x33ed7d2a,0xe72719c1,0x154c9ac2,0x61c6936,0xf477ea35,0xaa64d611,0x580f5512,0x4b5fa6e6,0xb93425e5,0x6dfe410e,0x9f95c20d,0x8cc531f9,0x7eaeb2fa,0x30e349b1,0xc288cab2,0xd1d83946,0x23b3ba45,0xf779deae,0x5125dad,0x1642ae59,0xe4292d5a,0xba3a117e,0x4851927d,0x5b016189,0xa96ae28a,0x7da08661,0x8fcb0562,0x9c9bf696,0x6ef07595,0x417b1dbc,0xb3109ebf,0xa0406d4b,0x522bee48,0x86e18aa3,0x748a09a0,0x67dafa54,0x95b17957,0xcba24573,0x39c9c670,0x2a993584,0xd8f2b687,0xc38d26c,0xfe53516f,0xed03a29b,0x1f682198,0x5125dad3,0xa34e59d0,0xb01eaa24,0x42752927,0x96bf4dcc,0x64d4cecf,0x77843d3b,0x85efbe38,0xdbfc821c,0x2997011f,0x3ac7f2eb,0xc8ac71e8,0x1c661503,0xee0d9600,0xfd5d65f4,0xf36e6f7,0x61c69362,0x93ad1061,0x80fde395,0x72966096,0xa65c047d,0x5437877e,0x4767748a,0xb50cf789,0xeb1fcbad,0x197448ae,0xa24bb5a,0xf84f3859,0x2c855cb2,0xdeeedfb1,0xcdbe2c45,0x3fd5af46,0x7198540d,0x83f3d70e,0x90a324fa,0x62c8a7f9,0xb602c312,0x44694011,0x5739b3e5,0xa55230e6,0xfb410cc2,0x92a8fc1,0x1a7a7c35,0xe811ff36,0x3cdb9bdd,0xceb018de,0xdde0eb2a,0x2f8b6829,0x82f63b78,0x709db87b,0x63cd4b8f,0x91a6c88c,0x456cac67,0xb7072f64,0xa457dc90,0x563c5f93,0x82f63b7,0xfa44e0b4,0xe9141340,0x1b7f9043,0xcfb5f4a8,0x3dde77ab,0x2e8e845f,0xdce5075c,0x92a8fc17,0x60c37f14,0x73938ce0,0x81f80fe3,0x55326b08,0xa759e80b,0xb4091bff,0x466298fc,0x1871a4d8,0xea1a27db,0xf94ad42f,0xb21572c,0xdfeb33c7,0x2d80b0c4,0x3ed04330,0xccbbc033,0xa24bb5a6,0x502036a5,0x4370c551,0xb11b4652,0x65d122b9,0x97baa1ba,0x84ea524e,0x7681d14d,0x2892ed69,0xdaf96e6a,0xc9a99d9e,0x3bc21e9d,0xef087a76,0x1d63f975,0xe330a81,0xfc588982,0xb21572c9,0x407ef1ca,0x532e023e,0xa145813d,0x758fe5d6,0x87e466d5,0x94b49521,0x66df1622,0x38cc2a06,0xcaa7a905,0xd9f75af1,0x2b9cd9f2,0xff56bd19,0xd3d3e1a,0x1e6dcdee,0xec064eed,0xc38d26c4,0x31e6a5c7,0x22b65633,0xd0ddd530,0x417b1db,0xf67c32d8,0xe52cc12c,0x1747422f,0x49547e0b,0xbb3ffd08,0xa86f0efc,0x5a048dff,0x8ecee914,0x7ca56a17,0x6ff599e3,0x9d9e1ae0,0xd3d3e1ab,0x21b862a8,0x32e8915c,0xc083125f,0x144976b4,0xe622f5b7,0xf5720643,0x7198540,0x590ab964,0xab613a67,0xb831c993,0x4a5a4a90,0x9e902e7b,0x6cfbad78,0x7fab5e8c,0x8dc0dd8f,0xe330a81a,0x115b2b19,0x20bd8ed,0xf0605bee,0x24aa3f05,0xd6c1bc06,0xc5914ff2,0x37faccf1,0x69e9f0d5,0x9b8273d6,0x88d28022,0x7ab90321,0xae7367ca,0x5c18e4c9,0x4f48173d,0xbd23943e,0xf36e6f75,0x105ec76,0x12551f82,0xe03e9c81,0x34f4f86a,0xc69f7b69,0xd5cf889d,0x27a40b9e,0x79b737ba,0x8bdcb4b9,0x988c474d,0x6ae7c44e,0xbe2da0a5,0x4c4623a6,0x5f16d052,0xad7d5351,0x0,0x13a29877,0x274530ee,0x34e7a899,0x4e8a61dc,0x5d28f9ab,0x69cf5132,0x7a6dc945,0x9d14c3b8,0x8eb65bcf,0xba51f356,0xa9f36b21,0xd39ea264,0xc03c3a13,0xf4db928a,0xe7790afd,0x3fc5f181,0x2c6769f6,0x1880c16f,0xb225918,0x714f905d,0x62ed082a,0x560aa0b3,0x45a838c4,0xa2d13239,0xb173aa4e,0x859402d7,0x96369aa0,0xec5b53e5,0xfff9cb92,0xcb1e630b,0xd8bcfb7c,0x7f8be302,0x6c297b75,0x58ced3ec,0x4b6c4b9b,0x310182de,0x22a31aa9,0x1644b230,0x5e62a47,0xe29f20ba,0xf13db8cd,0xc5da1054,0xd6788823,0xac154166,0xbfb7d911,0x8b507188,0x98f2e9ff,0x404e1283,0x53ec8af4,0x670b226d,0x74a9ba1a,0xec4735f,0x1d66eb28,0x298143b1,0x3a23dbc6,0xdd5ad13b,0xcef8494c,0xfa1fe1d5,0xe9bd79a2,0x93d0b0e7,0x80722890,0xb4958009,0xa737187e,0xff17c604,0xecb55e73,0xd852f6ea,0xcbf06e9d,0xb19da7d8,0xa23f3faf,0x96d89736,0x857a0f41,0x620305bc,0x71a19dcb,0x45463552,0x56e4ad25,0x2c896460,0x3f2bfc17,0xbcc548e,0x186eccf9,0xc0d23785,0xd370aff2,0xe797076b,0xf4359f1c,0x8e585659,0x9dface2e,0xa91d66b7,0xbabffec0,0x5dc6f43d,0x4e646c4a,0x7a83c4d3,0x69215ca4,0x134c95e1,0xee0d96,0x3409a50f,0x27ab3d78,0x809c2506,0x933ebd71,0xa7d915e8,0xb47b8d9f,0xce1644da,0xddb4dcad,0xe9537434,0xfaf1ec43,0x1d88e6be,0xe2a7ec9,0x3acdd650,0x296f4e27,0x53028762,0x40a01f15,0x7447b78c,0x67e52ffb,0xbf59d487,0xacfb4cf0,0x981ce469,0x8bbe7c1e,0xf1d3b55b,0xe2712d2c,0xd69685b5,0xc5341dc2,0x224d173f,0x31ef8f48,0x50827d1,0x16aabfa6,0x6cc776e3,0x7f65ee94,0x4b82460d,0x5820de7a,0xfbc3faf9,0xe861628e,0xdc86ca17,0xcf245260,0xb5499b25,0xa6eb0352,0x920cabcb,0x81ae33bc,0x66d73941,0x7575a136,0x419209af,0x523091d8,0x285d589d,0x3bffc0ea,0xf186873,0x1cbaf004,0xc4060b78,0xd7a4930f,0xe3433b96,0xf0e1a3e1,0x8a8c6aa4,0x992ef2d3,0xadc95a4a,0xbe6bc23d,0x5912c8c0,0x4ab050b7,0x7e57f82e,0x6df56059,0x1798a91c,0x43a316b,0x30dd99f2,0x237f0185,0x844819fb,0x97ea818c,0xa30d2915,0xb0afb162,0xcac27827,0xd960e050,0xed8748c9,0xfe25d0be,0x195cda43,0xafe4234,0x3e19eaad,0x2dbb72da,0x57d6bb9f,0x447423e8,0x70938b71,0x63311306,0xbb8de87a,0xa82f700d,0x9cc8d894,0x8f6a40e3,0xf50789a6,0xe6a511d1,0xd242b948,0xc1e0213f,0x26992bc2,0x353bb3b5,0x1dc1b2c,0x127e835b,0x68134a1e,0x7bb1d269,0x4f567af0,0x5cf4e287,0x4d43cfd,0x1776a48a,0x23910c13,0x30339464,0x4a5e5d21,0x59fcc556,0x6d1b6dcf,0x7eb9f5b8,0x99c0ff45,0x8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4f3b6143,0x6071c64,0xdd439b0d,0x947fe62a,0x6e26e32e,0x271a9e09,0xfc5e1960,0xb5626447,0x89766c2d,0xc04a110a,0x1b0e9663,0x5232eb44,0xa86bee40,0xe1579367,0x3a13140e,0x732f6929,0xcb4d68f7,0x827115d0,0x593592b9,0x1009ef9e,0xea50ea9a,0xa36c97bd,0x782810d4,0x31146df3,0x1a00cb32,0x533cb615,0x8878317c,0xc1444c5b,0x3b1d495f,0x72213478,0xa965b311,0xe059ce36,0x583bcfe8,0x1107b2cf,0xca4335a6,0x837f4881,0x79264d85,0x301a30a2,0xeb5eb7cb,0xa262caec,0x9e76c286,0xd74abfa1,0xc0e38c8,0x453245ef,0xbf6b40eb,0xf6573dcc,0x2d13baa5,0x642fc782,0xdc4dc65c,0x9571bb7b,0x4e353c12,0x7094135,0xfd504431,0xb46c3916,0x6f28be7f,0x2614c358,0x1700aeab,0x5e3cd38c,0x857854e5,0xcc4429c2,0x361d2cc6,0x7f2151e1,0xa465d688,0xed59abaf,0x553baa71,0x1c07d756,0xc743503f,0x8e7f2d18,0x7426281c,0x3d1a553b,0xe65ed252,0xaf62af75,0x9376a71f,0xda4ada38,0x10e5d51,0x48322076,0xb26b2572,0xfb575855,0x2013df3c,0x692fa21b,0xd14da3c5,0x9871dee2,0x4335598b,0xa0924ac,0xf05021a8,0xb96c5c8f,0x6228dbe6,0x2b14a6c1,0x34019664,0x7d3deb43,0xa6796c2a,0xef45110d,0x151c1409,0x5c20692e,0x8764ee47,0xce589360,0x763a92be,0x3f06ef99,0xe44268f0,0xad7e15d7,0x572710d3,0x1e1b6df4,0xc55fea9d,0x8c6397ba,0xb0779fd0,0xf94be2f7,0x220f659e,0x6b3318b9,0x916a1dbd,0xd856609a,0x312e7f3,0x4a2e9ad4,0xf24c9b0a,0xbb70e62d,0x60346144,0x29081c63,0xd3511967,0x9a6d6440,0x4129e329,0x8159e0e,0x3901f3fd,0x703d8eda,0xab7909b3,0xe2457494,0x181c7190,0x51200cb7,0x8a648bde,0xc358f6f9,0x7b3af727,0x32068a00,0xe9420d69,0xa07e704e,0x5a27754a,0x131b086d,0xc85f8f04,0x8163f223,0xbd77fa49,0xf44b876e,0x2f0f0007,0x66337d20,0x9c6a7824,0xd5560503,0xe12826a,0x472eff4d,0xff4cfe93,0xb67083b4,0x6d3404dd,0x240879fa,0xde517cfe,0x976d01d9,0x4c2986b0,0x515fb97,0x2e015d56,0x673d2071,0xbc79a718,0xf545da3f,0xf1cdf3b,0x4620a21c,0x9d642575,0xd4585852,0x6c3a598c,0x250624ab,0xfe42a3c2,0xb77edee5,0x4d27dbe1,0x41ba6c6,0xdf5f21af,0x96635c88,0xaa7754e2,0xe34b29c5,0x380faeac,0x7133d38b,0x8b6ad68f,0xc256aba8,0x19122cc1,0x502e51e6,0xe84c5038,0xa1702d1f,0x7a34aa76,0x3308d751,0xc951d255,0x806daf72,0x5b29281b,0x1215553c,0x230138cf,0x6a3d45e8,0xb179c281,0xf845bfa6,0x21cbaa2,0x4b20c785,0x906440ec,0xd9583dcb,0x613a3c15,0x28064132,0xf342c65b,0xba7ebb7c,0x4027be78,0x91bc35f,0xd25f4436,0x9b633911,0xa777317b,0xee4b4c5c,0x350fcb35,0x7c33b612,0x866ab316,0xcf56ce31,0x14124958,0x5d2e347f,0xe54c35a1,0xac704886,0x7734cfef,0x3e08b2c8,0xc451b7cc,0x8d6dcaeb,0x56294d82,0x1f1530a5,0};
unsigned int str2[10]={0x95F553DC,0x1D3B4E23,0x0B913414,0x6B9E4582,0x35E7621B,0x0FC86887F,0x26DEA0FB,0x87044CAA,0};
unsigned int v9,v12,v13,v14,v15,v16;
int v17,v18,v19,v20,v4;
long long int v7;
int main(void){
    for(register int round=0;round<8;round++)
        for(register int i1=0;i1<=15;i1++)
            for(register int i2=0;i2<=15;i2++)
                for(register int i3=0;i3<=15;i3++)
                    for(register int i4=0;i4<=15;i4++){
                        v9=-1;
                        char input[5]={c[i1],c[i2],c[i3],c[i4],0};
                        for(register int i=0;i<4;i++)
                            v9=str1[v7+(~(input[i]&v9)&(input[i]|v9)&0xFF)]^(v9>>8);
                        v12=(v9|~str2[v4])&~(v9&~str2[v4]);
                        v13=((((((((v12|(v12>>1))>>2)|v12|(v12>>1))>>4)|((v12|(v12>>1))>>2)|v12|(v12>>1))>>8)|((((v12|(v12>>1))>>2)|v12|(v12>>1))>>4)|((v12|(v12>>1))>>2)|v12|(v12>>1))>>16)|((((((v12|(v12>>1))>>2)|v12|(v12>>1))>>4)|((v12|(v12>>1))>>2)|v12|(v12>>1))>>8)|((((v12|(v12>>1))>>2)|v12|(v12>>1))>>4)|((v12|(v12>>1))>>2)|v12|(v12>>1);
                        v14=((v12&(v13|(v13>>1))&~(v13&(v13>>1)))>>1)|v12&(v13|(v13>>1))&~(v13&(v13>>1));
                        v15=(((v14>>2)|v14)>>4)|(v14>>2)|v14;
                        v16=(((v15>>8)|v15)>>16)|(v15>>8)|v15;
                        v17=(4*((2*v16)|v16))|(2*v16)|v16;
                        v18=(((16*v17)|v17)<<8)|(16*v17)|v17;
                        v19=(4*((2*(v13&1))|v13&1))|(2*(v13&1))|v13&1;
                        v20=(((16*v19)|v19)<<8)|(16*v19)|v19;
                        if(((v20|(v20<<16))&(~((v18<<16)|v18)|v18&1))==0){
                            v4+=1;
                            v7+=0x100;
                            printf("%s",input);
                            break;
                        };
                    };
    return 0;
};

[长城杯 2021 院校组]Just_cmp-re

抽象题别做。

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s1="flag{********************************}"
key=[0x00,0x00,0x00,0x00,0x00,0x37,0x07,0x0A,0x37,0x0A,0x08,0x0A,0x06,0x06,0x0B,0x38,0x07,0x0A,0x3B,0x08,0x38,0x0E,0x0F,0x3B,0x3A,0x0A,0x0B,0x06,0x09,0x07,0x3B,0x37,0x0D,0x0F,0x07,0x38,0x0F,0x00]
for i in range(len(s1)):
    print(chr(ord(s1[i])+key[i]),end='')

[suctf 2019]signin

密码题跑re里了…，经典RSA，$n$直接丢factordb里就能分解出来。

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from Crypto.Util.number import *
p=282164587459512124844245113950593348271
q=366669102002966856876605669837014229419
e=65537
c=0xad939ff59f6e70bcbfad406f2494993757eee98b91bc244184a377520d06fc35
phi=(p-1)*(q-1)
n=p*q
d=inverse(e,phi)
m=pow(c,d,n)
print(long_to_bytes(m))

[NCTF 2019]Our 16bit wars

16位程序阅读题，直接硬杠，写脚本爆破：

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enc=[0xC9,0x68,0x8A,0xC8,0x6F,0x07,0x06,0x0F,0x07,0xC6,0xEB,0x86,0x6E,0x6E,0x66,0xAD,0x4C,0x8D,0xAC,0xEB,0x26,0x6E,0xEB,0xCC,0xAE,0xCD,0x8C,0x86,0xAD,0x66,0xCD,0x8E,0x86,0x8D,0xAF]
for i in range(len(enc)):
    for j in range(0,0xff+1):
        if (j>>3)^(j<<5)&0xff==enc[i]:
            print(chr(j),end='')
            break

[LitCTF 2023]debase64

出现个Base64的加密函数，第一次见加密的。

主要加密流程为：

Base64加密几字节，注意大端序，截断在0xFF之前。

最后还有个字节不知道啥。

题目说末尾还有仨等于号。

MD5爆破：

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import hashlib,string
enc="Y0uReallyKn0wB4s"
for i in string.printable:
    tmp=enc+i+"==="
    if hashlib.md5(tmp.encode()).hexdigest()=='5a3ebb487ad0046e52db00570339aace':
        print(tmp)
        exit(0)

[网鼎杯 2020 青龙组]jocker

最后一步就很怪，代码是错的，必须猜…

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keys="hahahaha_do_you_find_me?"
enc=[0x0000000E,0x0000000D,0x00000009,0x00000006,0x00000013,0x00000005,0x00000058,0x00000056,0x0000003E,0x00000006,0x0000000C,0x0000003C,0x0000001F,0x00000057,0x00000014,0x0000006B,0x00000057,0x00000059,0x0000000D]
for i in range(len(enc)):
    print(chr(ord(keys[i])^enc[i]),end='')
key2=ord(":")^ord("}")
enc2="%tp&:"
for i in range(len(enc2)):
    print(chr(ord(enc2[i])^key2),end='')

[HDCTF 2023]easy_asm

16位程序逆向，只能看汇编。

汇编指令lodsb指的是将si寄存器指向地址内容一个一个字节读入al寄存器。与stosb指令配合使用，作用是将al寄存器内容放回si寄存器指向地址，相当于一个循环。

主要加密逻辑就是与寄s存器cl=0x10异或而已。

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enc=list("XTSDVkZecdOqOu#ciOqC}m".encode())
for i in range(len(enc)):
    enc[i]^=0x10
    print(chr(enc[i]),end='')

[HGAME 2023 week1]easyasm

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enc=[0x5b,0x54,0x52,0x5e,0x56,0x48,0x44,0x56,0x5f,0x50,0x3,0x5e,0x56,0x6c,0x47,0x3,0x6c,0x41,0x56,0x6c,0x44,0x5c,0x41,0x2,0x57,0x12,0x4e]
for i in range(len(enc)):
    print(chr(enc[i]^0x33),end='')

[CISCN 2022 东北]easycpp

连蒙带猜。

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enc=[0x0A,0x0B,0x7D,0x2F,0x7F,0x67,0x65,0x30,0x63,0x60,0x37,0x3F,0x3C,0x3F,0x33,0x3A,0x3C,0x3B,0x35,0x3C,0x3E,0x6C,0x64,0x31,0x64,0x6C,0x3B,0x68,0x61,0x62,0x65,0x36,0x33,0x60,0x62,0x36,0x1C,0x7D]
for i in range(len(enc)-4,0,-1):
    enc[i+2]^=enc[i+3]
    enc[i+1]^=enc[i+2]
    enc[i]^=enc[i+1]
for i in range(len(enc)):
    print(chr(enc[i]),end='')

[网鼎杯 2022 青龙组]fakeshell

逻辑挺简单的，就是代码有点骚。

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enc=[0x0000004B,0x00000048,0x00000079,0x00000013,0x00000045,0x00000030,0x0000005C,0x00000049,0x0000005A,0x00000079,0x00000013,0x00000070,0x0000006D,0x00000078,0x00000013,0x0000006F,0x00000048,0x0000005D,0x00000064,0x00000064]
for i in range(len(enc)):
    print(chr(((enc[i]^0x50)-10)^0x66),end='')