ISCC2024個人挑戰賽WP-迷失之門

（非官方解，以下內容均互聯網收集的信息和個人思路，僅供學習參考）

迷失之門

方法一：

IDA看一下 check函數邏輯

進入到check2函數

R鍵將ascii碼轉字符，寫出逆向腳本

#include <stdio.h>

#include <string.h>

int main() {

??? char cmp[] = "FSBBhKfIDirfZlEfVbaShZyKom";

??? char v16[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";

??? char v10[] = "abcdefghijklmnopqrstuvwxyz";

??? char v4[] = "0123456789+/-=!#&*()?;:*^%";

??? char v3[] = "DABBZXQESVFRWNGTHYJUMKIOLPC";

??? int len = strlen(cmp);

??? int v22;

??? char v1;

??? for (int i = 0; i < len; i++) {

??????? for (int j = 32; j < 127; j++) {

??????????? if (j - v3[i] <= 0) {

??????????????? continue;

??????????? } else {

??????????????? v22 = j - v3[i];

??????????????? if (v22 > 25) {

??????????????????? if (v22 > 51) {

??????????????????????? v1 = v4[v22 - 52];

??????????????????? } else {

??????????????????????? v1 = v10[v22 - 26];

??????????????????? }

??????????????????? if (v1 == cmp[i]) {

??????????????????????? printf("%c", j);

??????????????????????? break;

??????????????????? }

??????????????? } else {

??????????????????? if (v16[v22] == cmp[i]) {

??????????????????????? printf("%c", j);

??????????????????????? break;

??????????????????? }

??????????????? }

??????????? }

??????? }

??? }

??? return 0;

}

方法二:

打開ida 然后shift+F2

enc=''.join([chr(get_wide_byte(0x140001464+i*(0x77-0x64))) for i in range(20)]+[chr(get_wide_byte(0x1400015e0+i*15)) for i in range(7)])

key=[i for i in b"DABBZXQESVFRWNGTHYJUMKIOLPC"]

print([i+51 for i in key])

index=[]

for i in enc:

??? if i in "ABCDEFGHIJKLMNOPQRSTUVWXYZ":

??????? index.append("ABCDEFGHIJKLMNOPQRSTUVWXYZ".index(i))

??? elif i in "abcdefghijklmnopqrstuvwxyz":

??????? index.append("abcdefghijklmnopqrstuvwxyz".index(i)+0x1a)

??? elif i in "0123456789+/-=!#&*()?;:*^%":

??????? index.append("0123456789+/-=!#&*()?;:*^%".index(i)+0x34)

??? else:

??????? print("wrong")

print(index)

for i in range(len(index)):

??? print(chr((key[i]+index[i])),end='')

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