Easy Encryption

点击此处获得更好的阅读体验

---

WriteUp来源

https://xz.aliyun.com/t/6912

题目考点

解题思路

.base64解密字符串：YXJ0cWtvZWhxcGtiaWh2

2.将固定字符串按照规则解密即可

字符串：**artqkoehqpkbihv**

解密如下：

#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;

char base64[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
char resl[] = "artqkoehqpkbihv";
//D0M3GVOMTOac


char Vignia(char a,int n,int* str1,int h)
{
    return (((int)a - 97 + str1[n+h*15]) % 26 + 97) <= 122
                ? (char)(((int)a - 97 + str1[n+h*15]) % 26 + 97) :
                (char)(((int)a - 97 + str1[n+h*15]) % 26 + 97 - 26);

}

int main()
{
    int len,i, j;
    int key1[1000];
    char flag[30];
    char tmp;
    len = strlen(resl);
    i = 0;

    while (base64[i] != '\0')
    {
        key1[i] = (int)abs(base64[i] - 97);//取绝对值
        i++;
    }

    for (int h = 4; h < len; h++)
    {
        for (j = 48; j < 126; j++)
        {
            for (i = 0; i < 4; i++)
            {
                if (Vignia((char)j, h, key1, i) != resl[h])
                    break;
                flag[h] = (char)j;
            }

        }
    }

    for (int h = 0; h < 4; h++)
    {
        for (j = 48; j < 126; j++)
        {
            for (i = 0; i < 5; i++)
            {
                if (Vignia((char)j, h, key1, i) != resl[h])
                    break;
                flag[h] = (char)j;
            }

        }
    }

    flag[len] = 0x00;
    printf("%s", flag);
    printf("\n");
    return 0;

}

Flag

umpnineissogood