逆向签到

easy_magic

Just look at the main function directly

scanf()function but note that only one character is entered here（%c）就可以了

这里需要注意v5The value of is equal to the character we entered（v4）去异或0x77
而v10The first character of the array is equal to what we enteredv4

主算法

The main algorithm is very simple in two sentences ：
v5依次加上v7The value of , and then the result of each addition is givenv10数组
最后把v10数组md5Encrypted and then assigned tos1
最后比较s1和s2的md5值是否相同 If the same, the characters entered are correct

解题思路：
Although the questioner gave itmd5value but obviously to solve directlymd5Not the author's intention
But we need to askv10的值
Because the algorithm logic is very clear that it is simply lettingv5去累加v7And the result of each accumulation is assigned tov10
所以我们只需要知道v5The initial value of
因为比赛的flag格式为ctfshow{}所以可以猜测v5The initial value of must be c的asc码（99）

Just dig out the algorithm and decrypt it

#include <stdio.h>
#include<string.h>
int main()
{
    int  v4;
    char v5;
    char v7[32];
    char v10[104];
    v7[0] = 17;
    v7[1] = 0xF2;
    v7[2] = 13;
    v7[3] = 0xF5;
    v7[4] = 7;
    v7[5] = 8;
    v7[6] = 4;
    v7[7] = 0xBC;
    v7[8] = 65;
    v7[9] = 0xE7;
    v7[10] = 7;
    v7[11] = 6;
    v7[12] = 0xF5;
    v7[13] = 6;
    v7[14] = 0xF8;
    v7[15] = 10;
    v7[16] = 10;
    v7[17] = 0xEC;
    v7[18] = 9;
    v7[19] = 0xFD;
    v7[20] = 0xD;
    v7[21] = 0xF3;
    v7[22] = 24;
    v5 = 0;
    v4 = int('c') ^ 0x77;
    v5 = v4 ^ 0x77;
    v10[0] = 'c';
    for (int i = 0; i <= 22; ++i)
    {
        v5 += v7[i];
        v10[i + 1] = v5;
    }

    v10[24] = 0;

    for(int i = 0;i<=strlen(v10);i++)
    {
    printf("%c", v10[i]);
    }
}
//ctfshow{7x_flag_is_here}

This question is the idea given by the student Posted by the apprenticepython的脚本

v7 = [17,-14,13,-11,7,8,4,-68,65,-25,7,6,-11,6,-8,10,10,-20,9,-3,13,-13,24]
v4 = "c"
for i in range(len(v7)):

v5 = v4
v10 = ""
for i in range(len(v4)):
v10 += chr(ord(v5[i])+v7[i])

v4 = "c" + v10
print(v4)