蓝桥杯ctf2024 部分wp

数据分析

1. packet

密码破解

1. cc

逆向分析

1. 欢乐时光

XXTEA

#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(unsigned int* v, int n, unsigned int const key[4])
{
    unsigned int y, z, sum;
    unsigned p, rounds, e;
    if (n > 1)
    {
        rounds = 32;	//这里可以说是预定义值，n=2是rounds=32
        sum = 0;
        z = v[n - 1];
        do
        {
            sum += DELTA;
            e = (sum >> 2) & 3;
            for (p = 0; p < n - 1; p++)        //注意这里的p是从0~n-1
            {
                y = v[p + 1];
                z = v[p] += MX;
            }
            y = v[0];
            z = v[n - 1] += MX;        //这里的MX中传入的p=n-1
        } while (--rounds);
    }
    else if (n < -1)
    {
        n = -n;
        rounds = 415 / n + 114;
        sum = rounds * DELTA;
        y = v[0];
        do
        {
            e = (sum >> 2) & 3;
            for (p = n - 1; p > 0; p--)    //注意这里的p是从n-1~0,和上面是反过来的
            {
                z = v[p - 1];
                y = v[p] -= MX;
            }
            z = v[n - 1];
            y = v[0] -= MX;    //这里的MX中传入的 p=0
            sum -= DELTA;
        } while (--rounds);
    }
}


int main()
{
	int enc[] = { 0x480AC20C, 0xCE9037F2, 0x8C212018, 0x0E92A18D, 0xA4035274, 0x2473AAB1, 0xA9EFDB58, 0xA52CC5C8,
	0xE432CB51, 0xD04E9223, 0x6FD07093, 0 };
	int key[4] = {
	0x79696755, 0x67346F6C, 0x69231231, 0x5F674231
	};
    int n = 11;
	btea(enc,-n, key);
	printf("%s", enc);
	return 0;
}