re-upx

根据提示可知是upx加壳，脱壳后用IDA打开，惯例先按shift+F12,
得到：

可知需要输入验证，双击可执行文件也可验证猜想，双击“please input your flag: ”
跳转反汇编界面：

查看汇编代码和流程：

之后F5查看伪代码，

根据代码逻辑，我们可以逆推解密，v10的35个字符：

接下来逆推解密：
0x23, 0x2B, 0x27, 0x36, 0x33, 0x3C, 0x3, 0x48, 0x64, 0xB, 0x1D, 0x76, 0x7B, 0x10, 0xB, 0x3A, 0x3F, 0x65, 0x76, 0x29,
0x15, 0x37, 0x1C, 0xA, 0x8, 0x21, 0x3E, 0x3C, 0x3D, 0x16, 0xB, 0x24, 0x29, 0x24, 0x56,0xa
0x29, 0x24, 0x56,0xa分别是41，36，86，\n.
最后的结果：moectf{Y0u_c4n_unp4ck_It_vvith_upx}

代码：

#include <iostream>
#include <vector>

int main() {
    std::vector<unsigned char> v10_bytes = {
        0x23, 0x2B, 0x27, 0x36, 0x33, 0x3C, 0x3, 0x48, 0x64, 0xB, 0x1D, 0x76, 0x7B, 0x10, 0xB, 0x3A, 0x3F, 0x65, 0x76, 0x29,
        0x15, 0x37, 0x1C, 0xA, 0x8, 0x21, 0x3E, 0x3C, 0x3D, 0x16, 0xB, 0x24, 0x29, 0x24, 0x56, 0xa
    };
    
    std::vector<unsigned char> flag(v10_bytes.begin(), v10_bytes.end());
    
    for (int i = 35; i > 0; i--) {
        flag[i - 1] = flag[i - 1] ^ 0x21 ^ flag[i];
    }
    
    for (auto c : flag) {
        std::cout << c;
    }
    std::cout << std::endl;
    
    return 0;
}