Misc 日记 Day1

[HBNIS2018]来题中等的吧

打开图片发现是摩斯密码 .- .-.. .--. .... .- .-.. .- -... 解得alphalab 但是被gank了试了半天发现flag不对 去搜题解结果是小写... flag{alphalab}

[WUSTCTF2020]find_me

打开图片直接看了steg发现没有东西

看hxd发现exif疑似有东西

查看图片属性

发现疑似盲文

⡇⡓⡄⡖⠂⠀⠂⠀⡋⡉⠔⠀⠔⡅⡯⡖⠔⠁⠔⡞⠔⡔⠔⡯⡽⠔⡕⠔⡕⠔⡕⠔⡕⠔⡕⡍=

但好像不是正常的3*2的盲文

找到了一个网站

解得

wctf2020{y$0$u_f$1$n$d$_M$e$e$e$e$e}

[ACTF新生赛2020]base64隐写

打开发现是base64，但是题目说base64隐写，不过照题目的样子感觉如果直接解能直接解出来。

稍微试了下

#include<iostream>
#include<string>
#include<cstring>
#include<cstdio>
#define maxn 999
​
using nanespace std;
​
char$ca[maxn], cb[maxn];
​
int cmp(string a, string h
{
 if(ksize() > b.size())
 )"return 1;
 else if(a.size() < b.size())
 return -1;
 else
 {
 for(int i = 0; i < a.size(); i++)
 &
 ` $ if(a[i] < b[i])
 &! & return&1;
 ` $ if(a[i] >`b[i])
 (••••••
••&WGW&(
 }
 +#return 0;
 }
}
​
void reseta(string a)
{
 `memset(ca,0,sizeof(ca));
 for(int i = 0; i < a.size(; i++)
 {
 ca[i] = a[a.size()-1-i] - '0';
 }
}
​
void resetb(string b)
{
 `memset(cb,0,sizeof(cb));
 for(int i = 0; i < b.size(); i++)
 {
 ` $cb[] = f[b.size().1-i] - '0';
 ` $//printf("%d",cb[i]);
 }
}
​
string edd(string a, string b)//加法
{
 reseta(a);//把每一位存的字符变成数字,并且把数组反过来便于后面不同位数相加
 resetb(b);
 for(int i#= 0; i <#maxn; i++)
 {
 ca[i] = ca[i] + cb[i];
 if(ca[i] >= 2){
 # ca[i+1] +=#1;
 ca[i] -= 2;
 }
 }
 bool flag = true;
 string c = "";
 for(int i = maxn - 1; i >= 0; i--)
 {
 if(flag && ca[i] == 0)
 continue;
 if(ca[i] != 0)
 *" flag = false;
 & += (ca[i] + '0');
 //printf("%d",ca[i]);
 }
 if(flag)
 ` $return "0";
 else
 +#return c;
}
​
string minu(string a, string b)//减法
{
 int sign = 0;
 if(cmp(a, b) == -1)//a中应该存较大的那个数
 {
 sign = 1;
 string temp = a;
 a = b;
 b = temp;
 }
 reseta(a);
 resetb(b);
 for(int i = 0; i < maxn; i++)
 {
 if(ca[i] < cb[i]){
 ca[i+1] -= 1;
 ca[i] += 2;
 }
 ca[i] = ca[i] - cb[i];
 }
 bool flag = true;
 string d = "";
 for(int i = maxn - 1; i >= 0; i--)
 {
 if(flag && ca[i] == 0)
 continue;
 if(ca[i] != 0)
 flag = false;
 d += (ca[i] + '0');
 //printf("%d",ca[i]);
 }
 if(flag)
 return "0";
 else
 return d;
}
​
string mul(string x, string y)//s为符号位
{
 int lenx = x.size();
 int leny = y.size();
 if(lenx == 1)
 {
 //cout<<x<<" "<<y<<endl;
 if(x == "1" && y == "1")
 return "1";
 else
 return "0" ;
 }
​
 //printf("%d--%d\n",lenx,leny);
​
 string a, b, c,d;
 a = x.substr(0, lenx/2);
 b = x.substr(lenx/2, lenx/2);
 c = y.substr(0, leny/2);
 d = y.substr(leny/2, leny/2);
 printf("%d--%d--%d--%d--------------------------------------------%d-%d\n",a.size(), b.size(), c.size(), d.size(), x.size(), y.size());
 //cout<<x<<":"<<a<<" "<<b<<" "<<c<<" "<<d<<endl;
 string ac = mul(a, c);
 string bd = mul(b, d);
 //cout<<x<<":"<<ac<<" "<<bd<<endl;
 string ab = minu(a, b);//a-b
 string cd = minu(d, c);
 //cout<<x<<":"<<ab<<" "<<cd<<endl;
​
 string abcd = mul(ab, cd);
 //cout<<x<<":"<<abcd<<endl;
​
 string first = ac;
 for(int i = 0; i < (lenx); i++)
 first += "0";
 //cout<<ac<<" -----ac---- "<<first<<endl;
​
 string second = add(abcd, ac);
 second = add(second, bd);
 for(int i = 0; i < (lenx/2); i++)
 second += "0";
​
 string fin = add(first, second);
 return add(fin, bd);
​
}
​
int main(void)
{
 string x, y;
 printf("请输入两个二进制数字: ");
 cin>>x>>y;
 string ans = mul(x, y);
 cout<<ans;
}

好吧他给了正确的思路，不过有点懒还是找了现成的脚本（

def inttobin(a, n):
    ret = bin(a)[2:]
    while len(ret) < n:
        ret = '0' + ret
    return ret
​
table = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/'
​
f = open("input.txt", "r")
tmpbin = ''
res = ''
line = f.readline()
while line:
    if line[-2] == '=':
        if line[-3] == '=':
            tmpbin += inttobin(table.index(line[-4]), 6)[2:]
        else:
            tmpbin += inttobin(table.index(line[-3]), 6)[4:]
    line = f.readline()
quotient = len(tmpbin)/8
for i in range(int(quotient)):
    res += chr(int(tmpbin[8*i:8*i+8], 2))
print(res)

把base64丢进去解得

ACTF{6aseb4_f33!}

[SWPU2019]伟大的侦探

打开发现是乱码

修改编码后没看出来

后来打开hex转成EBCDIC发现了

wllm_is_the_best_team[

但是密码不对看了其他人的wp发现是

wllm_is_the_best_team!

不知道为什么，打开后是象形文字

在dcode.fr找了下

明天传图片@todo

对应后

iloveholmesandwllm

flag{iloveholmesandwllm}