funny game

misc

Google maze

f12 can slove

Sequences

show us the python script

sequence = [0, 1, 3, 6, 2]
numbers = [774, 5842, 4294, 9433, 5232, 1871, 3292, 4213, 2752, 5521, 4581, 3829, 2812, 2244, 3312, 7059, 2959, 8297, 7497, 1272, 2288, 160, 594, 158, 4833, 7140, 5352, 142, 3018, 9468, 5625, 4848, 2051, 3008, 4298, 6250, 144]
validPass = "ABCFEDFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz123457890{}_"

def gensequence():
    for i in range(10000):
        # TODO write sequence generator for one of the most famous sequences and add it to sequence
        pass
    
gensequence()

for i in numbers:
    print(validPass[sequence[i]%len(validPass)], end="")

the hint is ：Try looking for a list of the most famous sequences

wow，when i search 0 1 3 6 2 ，i found this link

Recaman’s sequence – GeeksforGeeks

ok,lets write script

exp.py

sequence = [0] * 10000
numbers = [774, 5842, 4294, 9433, 5232, 1871, 3292, 4213, 2752, 5521, 4581, 3829, 2812, 2244, 3312, 7059, 2959, 8297, 7497, 1272, 2288, 160, 594, 158, 4833, 7140, 5352, 142, 3018, 9468, 5625, 4848, 2051, 3008, 4298, 6250, 144]
validPass = "ABCFEDFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz123457890{}_"

def gensequence():
    for i in range(10000):
        curr = sequence[i-1] - i
        for j in range(0,i):
            if ((sequence[j]==curr) or curr< 0 ):
                curr = sequence[i-1] +i
                break
        sequence[i] = curr

    
gensequence()

for i in numbers:
    print(validPass[sequence[i]%len(validPass)], end="")

solve!

Blender Creation

010 editor open,At the bottom of the pic, you can saw a python script.

import bpy
from base64 import b64decode

bpy.ops.object.select_all(action="SELECT")
bpy.context.scene.cursor.location = (0, 0, 0)
bpy.ops.object.delete(use_global=False, confirm=False)
bpy.ops.object.camera_add(
    enter_editmode=False,
    align="VIEW",
    location=(0, 0, 0),
    rotation=(1.1169, 0.01420, 0.8269),
    scale=(1, 1, 1),
)
bpy.ops.transform.translate(
    value=(4.58569, -4.20269, 3.1699),
    orient_type="GLOBAL",
    orient_matrix=((1, 0, 0), (0, 1, 0), (0, 0, 1)),
)
bpy.ops.mesh.primitive_monkey_add(
    size=2, enter_editmode=False, align="WORLD", location=(0, 0, 0), scale=(1, 1, 1)
)
u =  "__"[1:] + 'x_N3r0f__'[2:-2][::-1]
yrsth = "b" + str(len(bpy.context.object.data.polygons[0].vertices) - 3) + str(len(bpy.context.object.data.polygons[2].vertices) - 1)+ "NdR__" +  "Xp_R3"[1::-1] + chr(len(bpy.context.object.data.polygons[3].vertices) + int(42.5*2)) + "YtH0"[1:-1] + str(len(bpy.context.object.data.polygons[5].vertices) ** 0 - 1) + "N_4ND" + u
bpy.ops.object.light_add(
    type="POINT", radius=1, align="WORLD", location=(0, -2.5, 0), scale=(1, 1, 1)
)
x = (
    int(
        0x0000000000000080
        + 0x0000000000000240
        + 0x0000000000000840
        - 0x0000000000000AC0
    )
    - 2
)
bpy.context.object.data.energy = 50
bpy.context.object.data.color = (1, 1, 0.69420)

a = (
    str(
        b64decode(
            (
                chr(x - 0x000000000000001)
                + chr(x - 0x000000000000001 * 10 // 2 + 6 - 6 % 4)
                + ("wemR3YhNmY")
            )[::-1]
        )
    )[:-1]
    + yrsth
    + "}"
)[2:]
bpy.data.curves.new(type="FONT", name="Font Curve").body = a
font_obj = bpy.data.objects.new(
    name="Font Object", object_data=bpy.data.curves["Font Curve"]
)
bpy.context.scene.collection.objects.link(font_obj)
font_obj.location = (0, 0.8, 1.2)
font_obj.rotation_euler = (90, 0, 0)
font_obj.scale = (0.75, 0.75, 0.75)

so,the a is the final flag,just print him is ok

print(a) can solve it .

foren

Broken Image

rename to png can solve

My New Friend

zsteg can solve ,ezgame

Superglue

funny game,

first,open by 010 editor,you can see jpg file header

You can view the three internal file types by using the binwalk command

so,we got there files,a png,jpg,and gif,

seem like a paragraph is missing,when we saw chall file then I looked at the bottom of the file again, I saw a WebP-style text, so I extracted it to get the last text.

final flag is this.

.bcapng

chall is give us width and height,so lets write python can solve it

from re import L
from PIL import Image
img = Image.new('RGB', (65, 79), (255, 255, 255))
w = 65
h = 79
bca = ''#01code
i = 0
for y in range(79):    
    for x in range(65):
        if bca[i] == '1':
            img.putpixel((x,y),(255,255,255))
            i+=1
        else:
            img.putpixel((x,y),(0,0,0))
            i+=1

img.show()