[GUET-CTF2019]encrypt 题解

#include<idc.idc>
static main(){
    Message("\nStart:-------------------------------------------------------------------\n");
    auto addr=0x602080;  
    auto i=0;
    auto length=51  ;
    for(i=addr;i<=addr+length;i++){
            Message("0x%02x,",Byte(i));
    }
    Message("End:-----------------------------------------------------------------------\n");
}

#include <stdio.h>
#include <string.h>
int main() {
    char data[] = {0x5a, 0x60, 0x54, 0x7A, 0x7A, 0x54, 0x72, 0x44, 0x7C, 0x66, 0x51, 0x50, 0x5B, 0x5F, 0x56, 0x56, 0x4C, 0x7C, 0x79, 0x6E, 0x65, 0x55, 0x52, 0x79, 0x55, 0x6D, 0x46, 0x6B, 0x6C, 0x56, 0x4A, 0x67, 0x4C, 0x61, 0x73, 0x4A, 0x72, 0x6F, 0x5A, 0x70, 0x48, 0x52, 0x78, 0x49, 0x55, 0x6C, 0x48, 0x5C, 0x76, 0x5A, 0x45, 0x3D};
    char flag[255];
    memset(flag, 0, sizeof(flag));
    int j = 0;
    char l[] = {0x10, 0x59, 0x9C, 0x92, 0x06, 0x22, 0xCF, 0xA5, 0x72, 0x1E, 0x45, 0x6A, 0x06, 0xCB, 0x08, 0xC3, 0xE4, 0x49, 0x5A, 0x63, 0x0C, 0xDF, 0xF6, 0x5F, 0x08, 0x28, 0xBD, 0xE2, 0x10, 0x15, 0x1F, 0x6E, 0xAA, 0x5A, 0xCA, 0xEC, 0x80, 0xAF, 0x9B, 0x16, 0xBB, 0x3D, 0x13, 0x2F, 0x6A, 0xA4, 0xC7, 0x2E, 0xBC, 0x4B, 0x60, 0x9A, 0xAF, 0xE9, 0xCE, 0xDA, 0x67, 0x39, 0xBA, 0x3B, 0x85, 0xEB, 0xD2, 0x6B, 0xAB, 0x06, 0x6B, 0x10, 0x57, 0x2C, 0x88, 0x70, 0xF7, 0x4F, 0xAA, 0x7F, 0x12, 0x47, 0xD6, 0xDE, 0x74, 0xB2, 0x1D, 0xA4, 0xD7, 0x76, 0x9A, 0xE0};
    for (int i = 0; i < sizeof(data); i += 4) {
        flag[strlen(flag)] = (((data[i] - 0x3D) & 0x3F) << 2) | (((data[i + 1] - 0x3D) & 0x30) >> 4);
        flag[strlen(flag)] = (((data[i + 1] - 0x3D) & 0x0F) << 4) | (((data[i + 2] - 0x3D) & 0x3C) >> 2);
        flag[strlen(flag)] = ((data[i + 3] - 0x3D) & 0x3F) | ((data[i + 2] - 0x3D) & 0x03) << 6;
    }
    for (int i = 0; i < strlen((const char *)flag); i++) {
       flag[i] ^= l[j++];
    }
    printf("%s", flag);
    return 0;
}