【湾区杯2025】Hardtest

本平台连接：[hardtest - Bugku CTF平台]

|> 这个题目拖进IDA里面发现整体很干净，没有混淆或者奇怪的加壳与花指令，可以尝试对伪代码进行算法还原

|>整体的分析转化为C语言差不多就这样（将就着看吧,反正不能运行）

>下面是伪代码，尝试直接进行分析

#include <bits/stdc++>
// Sbox查表 - 屯数据
char s_box[256] = {
    0x63, 0x7C, 0x77, 0x7B,叽里咕噜那一堆....
};
// 加密后的flag
char encrypted_flag[24] = {
    -105, -43, 96, 67, -76, 16, 67, 115, 15, -38, 67, -51, -45, -24, 115, 74, 
    -108, -61, -51, 113, -67, -36, -105, 26
};
// 循环左移函数
char rotate_left(char value, int shift) {
    return (value << shift) | (value >> (8 - shift));
}
// 循环右移函数
char rotate_right(char value, int shift) {
    return (value >> shift) | (value << (8 - shift));
}
// 在GF(257)有限域中计算模逆元
char modular_inverse(char a) {
    if (a == 0) return 0;
    short result = 1;
    short exponent = 255;
    short base = a;
    // 使用快速幂算法计算模逆（出题人还会算法哦）
    while (exponent) {
        if (exponent & 1) {
            result = (result * base) % 257;
        }
        base = (base * base) % 257;
        exponent >>= 1;
    }
    return (char)result;
}
// 主要字符变换函数
char transform_character(char input_char) {
    // 第一步：与0x5A异或后循环左移3位
    char step1 = rotate_left(input_char ^ 0x5A, 3);
    // 第二步：分离高4位和低4位
    char high_bits = (step1 >> 4) & 0xF;  // 高4位
    char low_bits = step1 & 0xF;          // 低4位
    // 第三步：在GF(16)上进行仿射变换
    char transformed_high = (3 * high_bits) & 0xF;  // 高4位乘以3
    char transformed_low = (5 * low_bits) & 0xF;    // 低4位乘以5
    // 第四步：重新组合高低位
    char combined = (transformed_high << 4) | transformed_low;
    // 第五步：在GF(257)上求模逆
    char inverse = modular_inverse(combined);
    // 第六步：使用S盒替换并循环右移2位
    return s_box[rotate_right(inverse, 2)];
}
// 预处理输入字符串
void preprocess_input(const char* input, char* output) {
    int length = strlen(input);
    for (int i = 0; i < length; ++i) {
        // 对每个字符进行循环左移，移位量由位置决定
        output[i] = rotate_left(input[i], (i % 7) + 1);
    }
}
int main() {
    char random_number = rand() % 255 + 1;
    printf("input your number(1-255): ");
    int user_input;
    if (random_number == user_input) {
        //碰运气哦
        while (getchar() != '\n');
        printf("flag: ");
        char user_flag[100];
        fgets(user_flag, sizeof(user_flag), stdin);
        // 移除换行符
        user_flag[strcspn(user_flag, "\n")] = 0;
        int flag_length = strlen(user_flag);
        // 预处理输入
        char* preprocessed_flag = (char*)malloc(flag_length + 1);
        preprocess_input(user_flag, preprocessed_flag);
        // 对每个字符进行变换
        char* transformed_flag = (char*)malloc(flag_length + 1);
        for (int i = 0; i < flag_length; ++i) {
            transformed_flag[i] = transform_character(preprocessed_flag[i]);
        }
        // 与目标flag比较
        int is_correct = 1;
        for (int i = 0; i < flag_length; ++i) {
            if (transformed_flag[i] != encrypted_flag[i]) {
                is_correct = 0;
                break;
            }
        }
        (is_correct) ? printf("right\n"):printf("error\n");
}

然后整体的加密与解密流程>>

>>>加密：
原始字符 → 预处理 → 异或0x5A → 循环左移3 → 分离高低位 → GF(16)乘法 → 组合 → GF(257)模逆 → S盒 → 循环右移2 → 密文

>>>逆向解密：密文 → 循环左移2 → 逆S盒 → GF(257)模逆 → 分离高低位 → GF(16)逆乘法 → 组合 → 循环右移3 → 异或0x5A → 逆向预处理 → 原始字符

由此可以写出简单的解密脚本喵~ >>

byte_2020 = [
    99, 124, 119, 123, 242, 107, 111, 197, 48, 1, 103, 43, 254, 215, 171, 118,
    202, 130, 201, 125, 250, 89, 71, 240, 173, 212, 162, 175, 156, 164, 114, 192,
    183, 253, 147, 38, 54, 63, 247, 204, 52, 165, 229, 241, 113, 216, 49, 21,
    4, 199, 35, 195, 24, 150, 5, 154, 7, 18, 128, 226, 235, 39, 178, 117,
    9, 131, 44, 26, 27, 110, 90, 160, 82, 59, 214, 179, 41, 227, 47, 132,
    83, 209, 0, 237, 32, 252, 177, 91, 106, 203, 190, 57, 74, 76, 88, 207,
    208, 239, 170, 251, 67, 77, 51, 133, 69, 249, 2, 127, 80, 60, 159, 168,
    81, 163, 64, 143, 146, 157, 56, 245, 188, 182, 218, 33, 16, 255, 243, 210,
    205, 12, 19, 236, 95, 151, 68, 23, 196, 167, 126, 61, 100, 93, 25, 115,
    96, 129, 79, 220, 34, 42, 144, 136, 70, 238, 184, 20, 222, 94, 11, 219,
    224, 50, 58, 10, 73, 6, 36, 92, 194, 211, 172, 98, 145, 149, 228, 121,
    231, 200, 55, 109, 141, 213, 78, 169, 108, 86, 244, 234, 101, 122, 174, 8,
    186, 120, 37, 46, 28, 166, 180, 198, 232, 221, 116, 31, 75, 189, 139, 138,
    112, 62, 181, 102, 72, 3, 246, 14, 97, 53, 87, 185, 134, 193, 29, 158,
    225, 248, 152, 17, 105, 217, 142, 148, 155, 30, 135, 233, 206, 85, 40, 223,
    140, 161, 137, 13, 191, 230, 66, 104, 65, 153, 45, 15, 176, 84, 187, 22
]

target = [b & 0xFF for b in [
    -105, -43, 96, 67, -76, 16, 67, 115, 15, -38, 67, -51, -45, -24, 115, 74,
    -108, -61, -51, 113, -67, -36, -105, 26
]]
# 计算逆元
def mod_inv(a, m=257):
    for i in range(1, m):
        if (a * i) % m == 1:
            return i
    return 0
#循环左移
def leftroll(x, s):
    return ((x << s) | (x >> (8 - s))) & 0xFF
#循环右移
def rightroll(x, s):
    return ((x >> s) | (x << (8 - s))) & 0xFF
#sbox逆向
def sbox(val):
    idx = byte_2020.index(val)
    v3 = leftroll(idx, 2)
    inv_v3 = mod_inv(v3)
    if inv_v3 == 0:
        return 0
    for v1 in range(256):
        high = (3 * (v1 >> 4)) & 0xF
        low = (5 * (v1 & 0xF)) & 0xF
        test = (16 * high) | low
        if test == inv_v3:
            return rightroll(v1, 3) ^ 0x5A
    return 0
flag = []
for i, b in enumerate(target):
    orig = sbox(b)
    shift = (i % 7) + 1
    flag.append(chr(rightroll(orig, shift)))
print(''.join(flag))

然后运行获得flag >>

flag{Bl@st1ng_1s_a_g00d_Way!!}