TEA篇

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搞完分组加密再看tea真的是神清气爽

TEA

#include <stdio.h>
#include <stdint.h>
#include <stdlib.h>
int main() {
    uint32_t sum, i; 
    uint32_t delta = 0x114514;
    uint32_t enc[10] = {0x78C594AB,0x22813B59,0x472A3144,0xF255108A,0x45CFB34,0x3949EA0C,
	0xCB760968,0x1559C979,0xDEF9929D,0x71D1AAB};
    unsigned int k[4] = {0x11451419,0x19810114,0x51419198,0x10114514};
    for (int m=0;m<10;m+=2)
    {
        sum = delta*32;
        for (i = 0; i < 32; i++) { 
		    enc[m + 1] -= (k[3] + (enc[m] >> 5)) ^ (sum + enc[m]) ^ (k[2] + 16 * enc[m]);
		    enc[m] -= (k[1] + (enc[m + 1] >> 5)) ^ (sum + enc[m + 1]) ^ (*k + 16 * enc[m + 1]);
		    sum -= delta;
        } 
    }
    for (int i=0;i<10;i++)
    {
        for(int m=0;m<=3;m++)
        {
            printf("%c",(enc[i]>>(m*8))&0xff);
        }
    }
    system("\npause");
}

XTEA

#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
void decrypt(uint32_t v[2], uint32_t const key[4]) {
    unsigned int i;
    uint32_t v0 = v[0], v1 = v[1], delta = 0x9E3779B9, sum = delta * (-32);
    for (i=0; i < 32; i++) {
        v1 -= (((v0 << 4) ^ (v0 >> 5)) + v0) ^ (sum + key[(sum>>11) & 3]);
        sum += delta;
        v0 -= (((v1 << 4) ^ (v1 >> 5)) + v1) ^ (sum + key[sum & 3]);

    }
    v[0] = v0; v[1] = v1;
}

int main() {
    uint32_t const k[4] = {0x19f8,0x11be,0x991,0x3418};
    uint32_t enc[] = {0x8ccb2324,0x9a7741a,0xfb3c678d,0xf6083a79,0xf1cc241b,0x39fa59f2,0xf2abe1cc,0x17189f72};
    for(int i=6; i>=0; i-=1){  //倒着解密
        decrypt(&enc[i], k);
    }
    for (int i = 0; i < 8; i++) {  
        for (int m = 3; m>=0; m--) {
            printf("%c", (enc[i] >> (8 * m)) & 0xff);
        }
    }
    system("\npause");
    return 0;
}

XXTEA

#include <stdio.h>
#include <stdint.h>
#include <string.h>
#include <stdlib.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) {         
         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) {  
        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[10] = {0x5D624C34, 0x8629FEAD, 0x9D11379B, 0xFCD53211, 0x460F63CE,
        0xC5816E68, 0xFE5300AD, 0x0A0015EE, 0x9806DBBB, 0xEF4A2648};
    uint32_t k[4] = {0x63656F6D, 0x21216674, 0x12345678, 0x9ABCDEF0};
    int n = 10; 
    btea(v, -n, k);
    for (int i = 0; i < 10; i++) {
        printf("%c", v[i] & 0xff);
        printf("%c", v[i] >> 8 & 0xff);
        printf("%c", v[i] >> 16 & 0xff);
        printf("%c", v[i] >> 24 & 0xff);
    }
    printf("\n");
    system("pause");
    return 0;
}