TEA算法

TEA算法

1.加密核心函数

void EncryptTEA(unsigned int *firstChunk, unsigned int *secondChunk, unsigned int* key)
{
    unsigned int y = *firstChunk;
    unsigned int z = *secondChunk;
    unsigned int sum = 0;

    unsigned int delta = 0x9e3779b9;

    for (int i = 0; i < 8; i++)//8轮运算(需要对应下面的解密核心函数的轮数一样)
    {
        sum += delta;
        y += ((z << 4) + key[0]) ^ (z + sum) ^ ((z >> 5) + key[1]);
        z += ((y << 4) + key[2]) ^ (y + sum) ^ ((y >> 5) + key[3]);
    }

    *firstChunk = y;
    *secondChunk = z;
}

2.解密核心函数

注意项：sum的初值不一定是0，根据加密函数写出

void DecryptTEA(unsigned int *firstChunk, unsigned int *secondChunk, unsigned int* key)
{
    unsigned int  sum = 0;
    unsigned int  y = *firstChunk;
    unsigned int  z = *secondChunk;
    unsigned int  delta = 0x9e3779b9;

    sum = delta << 3; //32轮运算，所以是2的5次方；16轮运算，所以是2的4次方；8轮运算，所以是2的3次方

    for (int i = 0; i < 8; i++) //8轮运算
    {
        z -= (y << 4) + key[2] ^ y + sum ^ (y >> 5) + key[3];
        y -= (z << 4) + key[0] ^ z + sum ^ (z >> 5) + key[1];
        sum -= delta;
    }

    *firstChunk = y;
    *secondChunk = z;
}

XXTEA算法
HGAME2021_weak1_apacha

#include <stdio.h>
#include <stdint.h>
#define DELTA 0x9e3779b9
#define MX (((z>>5^y<<2) + (y>>3^z<<4)) ^ ((sum^y) + (key[(p&3)^e] ^ z)))

void btea(uint32_t *v, int n, uint32_t const key[4])
{
	uint32_t y, z, sum;
	unsigned p, rounds, e;
	if (n > 1)            /* Coding Part */
	{
		rounds = 6 + 52 / n;
		sum = 0;
		z = v[n - 1];
		do
		{
			sum += DELTA;
			e = (sum >> 2) & 3;
			for (p = 0; p < n - 1; p++)
			{
				y = v[p + 1];
				z = v[p] += MX;
			}
			y = v[0];
			z = v[n - 1] += MX;
		} while (--rounds);
	}
	else if (n < -1)      /* Decoding Part */
	{
		n = -n;
		rounds = 6 + 52 / n;
		sum = rounds * DELTA;
		y = v[0];
		do
		{
			e = (sum >> 2) & 3;
			for (p = n - 1; p > 0; p--)
			{
				z = v[p - 1];
				y = v[p] -= MX;
			}
			z = v[n - 1];
			y = v[0] -= MX;
			sum -= DELTA;
		} while (--rounds);
	}
}
int main()
{
	uint32_t v[35] = {0xe74eb323,0xb7a72836,0x59ca6fe2,0x967cc5c1,0xe7802674,0x3d2d54e6,0x8a9d0356,0x99dcc39c,0x7026d8ed,0x6a33fdad,0xf496550a,0x5c9c6f9e,0x1be5d04c,0x6723ae17,0x5270a5c2,0xac42130a,0x84be67b2,0x705cc779,0x5c513d98,0xfb36da2d,0x22179645,0x5ce3529d,0xd189e1fb,0xe85bd489,0x73c8d11f,0x54b5c196,0xb67cb490,0x2117e4ca,0x9de3f994,0x2f5aa1aa,0xa7e801fd,0xc30d6eab,0x1baddc9c,0x3453b04a,0x92a406f9};
	uint32_t const k[4] = {1,2,3,4};
	int n = -35; //n的绝对值表示v的长度，取正表示加密，取负表示解密
	// v为要加密的数据是两个32位无符号整数
	// k为加密解密密钥，为4个32位无符号整数，即密钥长度为128位;
	btea(v, n, k);
	for (int i = 0; i < 35; i++)
	{
		printf("%c", (char)v[i]);

	}
	return 0;
}
//hgame{l00ks_1ike_y0u_f0Und_th3_t34}

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