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;
}