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这是某团队的招新赛总体难度还是不大的，这里写一下RE一道题目的WP

ezre

一道简单题，总结就是换表base64，变异RC4

分析


很简单的逻辑


第一个函数对key进行处理，由于对key处理的各种参数都是明文的，所以我们可以在这里下断点获取处理之后的key。


第二个函数对flag操作，将flag与一个变换后的key进行操作，由于变换的key比较复杂，同时变换的key同样与我们输入的参数无关，只需要使用原来的c代码就可以获取



第三个函数对flag最后处理，有个很明显的码表，大概能猜到是base64，仔细看看也能确定是base64。

解题脚本


下断点拿到key的值

通过脚本拿到xor_key的值

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#include <iostream>

#include <stdint.h>
#define HIDWORD(x) ((uint32_t)(((uint64_t)(x) >> 32) & 0xFFFFFFFF))

int main() {
    int v6; // [rsp+24h] [rbp-14h]
    int v7; // [rsp+28h] [rbp-10h]
    unsigned __int64 i; // [rsp+30h] [rbp-8h]
    int v3;
    int v5;
    int len = 21;

    v6 = 0;
    v7 = 0;
    unsigned char key[128] = {
            0x54, 0x0D, 0x08, 0x60, 0x14, 0x2C, 0x41, 0x2A, 0x30, 0x2E, 0x1C, 0x66, 0x1B, 0x47, 0x32, 0x74,
            0x65, 0x05, 0x68, 0x7E, 0x23, 0x24, 0x52, 0x5C, 0x48, 0x71, 0x11, 0x21, 0x25, 0x04, 0x3E, 0x4D,
            0x5B, 0x4C, 0x17, 0x29, 0x78, 0x45, 0x00, 0x3C, 0x7B, 0x6B, 0x6A, 0x5A, 0x50, 0x61, 0x19, 0x15,
            0x73, 0x7D, 0x75, 0x43, 0x3D, 0x3A, 0x70, 0x16, 0x77, 0x0C, 0x67, 0x51, 0x6F, 0x03, 0x6D, 0x58,
            0x4E, 0x37, 0x12, 0x2D, 0x4A, 0x1A, 0x4F, 0x5F, 0x4B, 0x7C, 0x55, 0x0F, 0x1D, 0x0E, 0x31, 0x6E,
            0x79, 0x1E, 0x22, 0x36, 0x69, 0x7A, 0x28, 0x26, 0x53, 0x56, 0x0B, 0x63, 0x5E, 0x64, 0x72, 0x3B,
            0x5D, 0x0A, 0x42, 0x01, 0x2F, 0x13, 0x09, 0x46, 0x3F, 0x6C, 0x7F, 0x44, 0x1F, 0x34, 0x18, 0x57,
            0x20, 0x39, 0x38, 0x02, 0x76, 0x10, 0x59, 0x49, 0x07, 0x27, 0x40, 0x2B, 0x35, 0x33, 0x62,0x06
    };
    for ( i = 0LL; ; ++i )
    {
        if ( i >= len )
            break;
        v6 = (v6 + 1) % 128;
        v3 = *(v6 + key) + v7;                      // v3 = key[v6] + v7
        v7 = (((HIDWORD(v3) >> 25) + *(v6 + key) + v7) & 0x7F) - (HIDWORD(v3) >> 25);
        v5 = *(v6 + key);                           // v5 = key[v6]
        *(v6 + key) = *(v7 + key);                  // key[v6] = key[v7]
        *(key + v7) = v5;                           // key[v7] = v5
        printf("%d,",*(((*(v6 + key) + *(v7 + key)) & 0x7F) + key));// flag[i] ^= key[(key[v6] + key[v7]) & 0x7f]
    }     // key[v7] = v5
}
//105,80,32,18,9,17,58,107,45,32,87,66,88,25,63,99,80,56,12,95,94,

编写python脚本解密即可

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import base64
import string

key = [0x54, 0x0D, 0x08, 0x60, 0x14, 0x2C, 0x41, 0x2A, 0x30, 0x2E, 0x1C, 0x66, 0x1B, 0x47, 0x32, 0x74, 0x65, 0x05, 0x68, 0x7E, 0x23, 0x24, 0x52, 0x5C, 0x48, 0x71, 0x11, 0x21, 0x25, 0x04, 0x3E, 0x4D, 0x5B, 0x4C, 0x17, 0x29, 0x78, 0x45, 0x00, 0x3C, 0x7B, 0x6B, 0x6A, 0x5A, 0x50, 0x61, 0x19, 0x15, 0x73, 0x7D, 0x75, 0x43, 0x3D, 0x3A, 0x70, 0x16, 0x77, 0x0C, 0x67, 0x51, 0x6F, 0x03, 0x6D, 0x58, 0x4E, 0x37, 0x12, 0x2D, 0x4A, 0x1A, 0x4F, 0x5F, 0x4B, 0x7C, 0x55, 0x0F, 0x1D, 0x0E, 0x31, 0x6E, 0x79, 0x1E, 0x22, 0x36, 0x69, 0x7A, 0x28, 0x26, 0x53, 0x56, 0x0B, 0x63, 0x5E, 0x64, 0x72, 0x3B, 0x5D, 0x0A, 0x42, 0x01, 0x2F, 0x13, 0x09, 0x46, 0x3F, 0x6C, 0x7F, 0x44, 0x1F, 0x34, 0x18, 0x57, 0x20, 0x39, 0x38, 0x02, 0x76, 0x10, 0x59, 0x49, 0x07, 0x27, 0x40, 0x2B, 0x35, 0x33, 0x62, 0x06]

enc_flag = "3pn1Ek92hmAEg38EXMn99J9YBf8="
string1 = "0123456789XYZabcdefghijklABCDEFGHIJKLMNOPQRSTUVWmnopqrstuvwxyz+/="
string2 = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/="
enc_flag1 = base64.b64decode(enc_flag.translate(str.maketrans(string1,string2))) #string1码表替换string2

xor_key = [105,80,32,18,9,17,58,107,45,32,87,66,88,25,63,99,80,56,12,95,94,]
flag = ''

for i in range(1,20):
    flag += chr(enc_flag1[i]^xor_key[i])

print(flag)
//lag{Simple_rEvErse}

得到结果可能是数据缺失，得到结果即可，仔细看其实是RC4算法，过程中灵活运用来减少时间。

snail

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