一些脚本收集

PNG图片隐写改高宽的反CRC

import os
import binascii
import struct

crcbp = open("C:\\Users\\Smartisan\\Desktop\\e1.png", "rb").read()    #打开图片
for i in range(2000):
    for j in range(2000):
        data = crcbp[12:16] + \
            struct.pack('>i', i)+struct.pack('>i', j)+crcbp[24:29]
        crc32 = binascii.crc32(data) & 0xffffffff
        if(crc32 == 0xa8586b45):    #图片当前CRC，自行修改
            print(i, j)
            print('hex:', hex(i), hex(j))

直接“顺序”写CRC

89 50 4E 47 0D 0A 1A 0APNG文件头

00 00 00 0DIDCH文件头数据块

49 48 44 52IHDR文件头数据块，固定为IHDR字符

接下来四字为宽度，如上图中00 00 02 00，以像素为单位

接下来四字为高度，如上图中00 00 02 C2，以像素为单位

接下来五字节依次表示：

图像深度，如上图中08

颜色类型，如上图中06

压缩方法，如上图中00

滤波器方法，如上图中00

隔行扫描方法，如上图中00

接下来的四字为CRC32校验码，如上图中A8 58 6B 45

参考文献（选自）：PNG文件格式具体解释 - mengfanrong - 博客园 (cnblogs.com)

ANSI escape code

FILE = open("flag", 'wb')
def writeThings(b):
    FILE.write(b)
def MoveLeft(n=1):
    writeThings(b'\x1b[C' * n)
def MoveRight(n=1):
    writeThings(b'\x1b[D' * n)
FLAG = b' 12312344444444' 
import random
COUNT = 2022234
randpos = [random.randint(0, len(FLAG) - 1) for _ in range(COUNT)]
randchar = [random.randint(0x20, 0x7e) for _ in range(COUNT)]
for i in range(len(FLAG)):
    pos = COUNT
    for j in range(COUNT - 1, 0, -1):
        if (randpos[j] == i):
            pos = j
            break
    randchar[pos] = FLAG[i]
writeThings(b'\n')
POS = 0
for i in range(COUNT):
    if (randpos[i] > POS):
        MoveLeft(randpos[i] - POS)
    elif (randpos[i] < POS):
        MoveRight(POS - randpos[i])
    writeThings(bytearray([randchar[i]]))
    POS = randpos[i] + 1
writeThings(b'\n')

ANSI escape code 控制光标移动输出随机字符，输出量足够多然后找到flag
cmd type flag // Terminal cat flag // Powershell cat flag type flag // type效率很低，很慢

已知(e, n, c)的RSA

import gmpy2
import binascii
c=110674792674017748243232351185896019660434718342001686906527789876264976328686134101972125493938434992787002915562500475480693297360867681000092725583284616353543422388489208114545007138606543678040798651836027433383282177081034151589935024292017207209056829250152219183518400364871109559825679273502274955582
n=135127138348299757374196447062640858416920350098320099993115949719051354213545596643216739555453946196078110834726375475981791223069451364024181952818056802089567064926510294124594174478123216516600368334763849206942942824711531334239106807454086389211139153023662266125937481669520771879355089997671125020789
e=65537
p = 11239134987804993586763559028187245057652550219515201768644770733869088185320740938450178816138394844329723311433549899499795775655921261664087997097294813
q = 12022912661420941592569751731802639375088427463430162252113082619617837010913002515450223656942836378041122163833359097910935638423464006252814266959128953
phi = (p-1)*(q-1)
d = gmpy2.invert(e,phi)
m = gmpy2.powmod(c,d,n)

print(binascii.unhexlify(hex(m)[2:]))

p，q在factordb.com进行质因数分解