下列十六进制数据,两个字节一组,找到其规律并用公式表达出来 0x6A,0x0E, 0x9E,0x0E, 0xD4,0x0E, 0xFC,0x0E, 0x3B,0x0F, 0x7A,0x0F, 0xAE,0x0F, 0xF0,0x0F, 0x2D,0x10, 0x69,0x10, 0xB3,0x10, 0x03,0x11, 0x37,0x11, 0x7F,0x11, 0xCF,0x11, 0x1E,0x12, 0x4E,0x12, 0x94,0x12, 0xDE,0x12, 0x11,0x13, 0x48,0x13, 0x7E,0x13, 0xAB,0x13, 0xC1,0x13, 0xD8,0x13, 0xDC,0x13, 0xDF,0x13,
时间: 2023-09-17 20:14:35 浏览: 27
根据观察,可以发现每组两个字节的第二个字节都是固定的0x0E,接下来第一个字节的值每隔4个字节就会加1,直到0xF,然后再从0x0开始循环。因此,可以用如下公式来表达这个规律:
第n组的第一个字节值 = (n mod 4) * 0x11 + (n div 16) * 0x1
其中,n表示组的编号,从0开始计数;mod表示取模运算;div表示整除运算。
相关问题
python 十六进制流
在 Python 中,可以使用以下两种方式表示十六进制流:
1. 使用字符串表示十六进制流:
```python
hex_data = "4a 6f 68 6e 20 44 6f 65 20 6a 75 73 74 20 6c 69 6b 65 20 74 6f 20 70 72 6f 67 72 61 6d"
byte_data = bytes.fromhex(hex_data.replace(' ', ''))
```
2. 使用字节数组表示十六进制流:
```python
hex_data = [0x4a, 0x6f, 0x68, 0x6e, 0x20, 0x44, 0x6f, 0x65, 0x20, 0x6a, 0x75, 0x73, 0x74, 0x20, 0x6c, 0x69, 0x6b, 0x65, 0x20, 0x74, 0x6f, 0x20, 0x70, 0x72, 0x6f, 0x67, 0x72, 0x61, 0x6d]
byte_data = bytes(hex_data)
```
其中,`byte_data` 表示转换后的字节流。
int main() { unsigned int v19; const unsigned char sbox0[256] ={0x63,0x7c,0x77,0x7b,0xf2,0x6b,0x6f,0xc5,0x30,0x01,0x67,0x2b,0xfe,0xd7,0xab,0x76,0xca,0x82,0xc9,0x7d,0xfa,0x59,0x47,0xf0,0xad,0xd4,0xa2,0xaf,0x9c,0xa4,0x72,0xc0,0xb7,0xfd,0x93,0x26,0x36,0x3f,0xf7,0xcc,0x34,0xa5,0xe5,0xf1,0x71,0xd8,0x31,0x15,0x04,0xc7,0x23,0xc3,0x18,0x96,0x05,0x9a,0x07,0x12,0x80,0xe2,0xeb,0x27,0xb2,0x75,0x09,0x83,0x2c,0x1a,0x1b,0x6e,0x5a,0xa0,0x52,0x3b,0xd6,0xb3,0x29,0xe3,0x2f,0x84,0x53,0xd1,0x00,0xed,0x20,0xfc,0xb1,0x5b,0x6a,0xcb,0xbe,0x39,0x4a,0x4c,0x58,0xcf,0xd0,0xef,0xaa,0xfb,0x43,0x4d,0x33,0x85,0x45,0xf9,0x02,0x7f,0x50,0x3c,0x9f,0xa8,0x51,0xa3,0x40,0x8f,0x92,0x9d,0x38,0xf5,0xbc,0xb6,0xda,0x21,0x10,0xff,0xf3,0xd2,0xcd,0x0c,0x13,0xec,0x5f,0x97,0x44,0x17,0xc4,0xa7,0x7e,0x3d,0x64,0x5d,0x19,0x73,0x60,0x81,0x4f,0xdc,0x22,0x2a,0x90,0x88,0x46,0xee,0xb8,0x14,0xde,0x5e,0x0b,0xdb,0xe0,0x32,0x3a,0x0a,0x49,0x06,0x24,0x5c,0xc2,0xd3,0xac,0x62,0x91,0x95,0xe4,0x79,0xe7,0xc8,0x37,0x6d,0x8d,0xd5,0x4e,0xa9,0x6c,0x56,0xf4,0xea,0x65,0x7a,0xae,0x08,0xba,0x78,0x25,0x2e,0x1c,0xa6,0xb4,0xc6,0xe8,0xdd,0x74,0x1f,0x4b,0xbd,0x8b,0x8a,0x70,0x3e,0xb5,0x66,0x48,0x03,0xf6,0x0e,0x61,0x35,0x57,0xb9,0x86,0xc1,0x1d,0x9e,0xe1,0xf8,0x98,0x11,0x69,0xd9,0x8e,0x94,0x9b,0x1e,0x87,0xe9,0xce,0x55,0x28,0xdf,0x8c,0xa1,0x89,0x0d,0xbf,0xe6,0x42,0x68,0x41,0x99,0x2d,0x0f,0xb0,0x54,0xbb,0x16}; const char* a3="UK*@3oKpFlVVnadsTfdA"; int v7=16; memcpy(&v19, a3, v7); for (int j = 0; j != 16; ++j ) *((_BYTE *)&v19 + j) = sbox0[*((unsigned *)&v19 + j)]; return 0; }输出v19值
根据代码可以看出,将字符串"a3"的前16个字符拷贝到一个unsigned int类型的变量v19中,并对v19进行一系列的sbox0变换。但是由于sbox0变换是按照字节进行的,因此需要先将v19的字节序进行转换。最终输出v19的值即可。
由于题目中没有给出字符串"a3",因此无法得知v19的值。