人工智能原理复习3 Informed Search alt text Blind search vs. “warmer” “colder” Info about where goal is. Optimal solution, explore less of the tree. Search Heuristics A heuristic is: A function that estimat 2024-06-18 人工智能原理
人工智能原理复习2 ai.berkeley.edu(课件资料来源) Search Problems Uninformed Search Methods Depth-First Search Breadth-First Search Uniform-Cost Search Unified formalism leading to A* search (search guided by heuristic 2024-06-17 人工智能原理
人工智能原理复习1 What is artificial intelligence? A (Short) History of AI 1940-1950: Early days 1943: McCulloch & Pitts: Boolean circuit model of brain 1950: Turing's “Computing Machinery and Intelligence” 19 2024-06-17 人工智能原理
2024XCTF4/1-5/1 签到题 题目: 123456789101112131415161718192021222324252627282930313233343536373839404142from Crypto.Util.number import *from tqdm import *import gmpy2flag=b'XYCTF{uuid}'flag=bytes_to_ 2024-04-12 CTF
pyecharts绘制中国地图 pyecharts库 通过测试发现,安装0.1.9.4版本最为稳定 1pip install pyecharts==0.1.9.4 安装完成后,下面是一个简单的测试案例 12345678910111213141516171819202122232425262728293031323334353637383940414243444546from pyecharts import Mappro 2024-04-10 python
WP_2024NKCTF_CRYPTO_ezmath 题目 123456789101112131415161718192021222324from Crypto.Util.number import *from secret import flagm1, m2 = bytes_to_long(flag[:len(flag)//2]), bytes_to_long(flag[len(flag)//2:])p, q, r, s = [getStrong 2024-03-26 CTF
2024_VCTF_crypto_狂飙 题目 123456789101112131415161718192021import osfrom secrets import flagfrom Crypto.Util.number import *from Crypto.Cipher import AESm = 88007513702424243702066490849596817304827839547007641526433597788 2024-03-18 CTF #VCTF
gmpy2库的常用函数总结 gmpy2 12345678910import gmpy2gmpy2.mpz(n) # 初始化一个大整数gmpy2.mpfr(x) # 初始化一个高精度浮点数xd = gmpy2.invert(e, n) # 求逆元,de = 1 mod n 模反元素C = gmpy2.powmod(M, e, n) # 幂取模,结果是 2024-03-11 python
WP_2024HSCCTF_CRYPTO_STAR_CHASING_DIARY 题目 123456789101112131415161718192021222324252627282930313233343536"""小肖在学习的过程中,了解到了一种填充图片的算法,兴奋的她用它和RSA加密了偶像的照片,但是在传输的过程中丢失了一部分加密算法,你能帮她补全,并解出来偶像的照片吗?"""from PIL import 2024-03-10 CTF
WP_2024HSCCTF_CRYPTO_MIXED 题目 12345678910111213141516171819202122232425262728293031import libnumfrom flag import flagfrom Crypto.Util.number import *m = libnum.s2n(flag)e = 65537q = getPrime(1024)q1 = getPrime(1024)p = getPrim 2024-03-10 CTF
WP_2024HSCCTF_CRYPTO_SING_IN 题目 123456789101112131415161718192021from Crypto.Util.number import *from random import *from gmpy2 import *flag = 'HSCCTF{66666666666666666666666666666666666}'m = bytes_to_long(fl 2024-03-10 CTF