目录 Buuctf crypto 0x01传感器 提示是曼联,猜测为曼彻斯特密码 wp:https://www.xmsec.cc/manchester-encode/ cipher: 5555555595555A65556AA696AA6666666955 cipher='5555555595555A65556AA696AA6666666955' def iee(cipher): tmp='' for i in range(len(cipher)): a=bin(eval('0x'+cipher

转载请注明出处: http://www.cnblogs.com/fraud/          ——by fraud Zball in Tina Town Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 219    Accepted Submission(s): 144 Problem Description Tina Town i

YAPTCHA Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 875    Accepted Submission(s): 458 Problem Description The math department has been having problems lately. Due to immense amount of uns

传送门 题意: 给出自然数 n,计算出 Sn 的值,其中 [ x ]表示不大于 x 的最大整数. 题解: 根据威尔逊定理,如果 p 为素数,那么 (p-1)! ≡ -1(mod p),即 (p-1)! + 1 = p*q. 令 f(K) =  ①如果 3K+7 为素数,则 (3K+7-1)! ≡ -1(mod 3K+7),即 (3K+6)! = (3K+7)*q -1. 那么表达式 可化简为 [ (3K+7)*q / (3K+7) - 1 / (3K+7) ] = [ q - 1 / (3K+7

<题目链接> 题目大意: The task that is presented to anyone visiting the start page of the math department is as follows: given a natural n, compute  where [x] denotes the largest integer not greater than x. 给出 t 和n,t代表样例组数,根据给出的n算出上面表达式.(注意:[x]表示,不超过x的最大整数)

<题目链接> Zball in Tina Town Problem Description Tina Town is a friendly place. People there care about each other.Tina has a ball called zball. Zball is magic. It grows larger every day. On the first day, it becomes 1 time as large as its original siz

YAPTCHA Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1490    Accepted Submission(s): 811 Problem Description The math department has been having problems lately. Due to immense amount of uns

题意就是叫你求上述那个公式在不同N下的结果. 思路:很显然的将上述式子换下元另p=3k+7则有 Σ[(p-1)!+1/p-[(p-1)!/p]] 接下来用到一个威尔逊定理,如果p为素数则 ( p -1 )! ≡ -1 ( mod p )    即 (p-1)!+1  为 p的整数倍  因此不难发现[*]里面要么为0,要么为1,为1的情况就是p为素数的情况,然后统计k=1-n中 有多少个3*k+1素数就好了 #include <iostream> #include <cstdio>

参考博客 HDU-2973 题目 Problem Description The math department has been having problems lately. Due to immense amount of unsolicited automated programs which were crawling across their pages, they decided to put Yet-Another-Public-Turing-Test-to-Tell-Compu

The math department has been having problems lately. Due to immense amount of unsolicited automated programs which were crawling across their pages, they decided to put Yet-Another-Public-Turing-Test-to-Tell-Computers-and-Humans-Apart on their webpag

Problem Description Farmer John keeps a website called ‘FansBlog’ .Everyday , there are many people visited this blog.One day, he find the visits has reached P , which is a prime number.He thinks it is a interesting fact.And he remembers that the vis

题意 给定一个整数 $P$($10^9 \leq p\leq 1^{14}$),设其前一个质数为 $Q$,求 $Q!  \ \% P$. 分析 暴力...说不定好的板子能过. 根据威尔逊定理,如果 $p$ 为质数,则有 $(p-1)! \equiv p-1(mod \ p)$. $\displaystyle Q! = \frac{(P-1)!}{(Q+1)(Q+2)...(p-1)} \equiv  (p-1)*inv\ (mod \ P)$. 根据素数定理,$\displaystyle \pi

威尔逊定理 在初等数论中,威尔逊定理给出了判定一个自然数是否为素数的充分必要条件.即:当且仅当p为素数时:( p -1 )! ≡ -1 ( mod p ),但是由于阶乘是呈爆炸增长的,其结论对于实际操作意义不大. 充分性 如果“p”不是素数,那么它的正因数必然包含在整数1, 2, 3, 4, … ,p− 1 中,因此gcd((p− 1)!,p) > 1,所以我们不可能得到(p− 1)! ≡ −1 (modp).   必要性   若p是素数,取集合 A={1,2,3,...p -1}; 则A 构成

YAPTCHA Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1885    Accepted Submission(s): 971   Problem Description The math department has been having problems lately. Due to immense amount of u

题意:给定质数p,求q!模p的值,其中q为小于p的最大质数 1e9<=p<=1e14 思路:根据质数密度近似分布可以暴力找q并检查 找到q后根据威尔逊定理: 把q+1到p-1这一段的逆元移过去 #include<bits/stdc++.h> using namespace std; typedef long long ll; typedef unsigned int uint; typedef unsigned long long ull; typedef pair<int,

Fansblog 题目传送门 解题思路 Q! % P = (P-1)!/(P-1)...(Q-1) % P. 因为P是质数,根据威尔逊定理,(P-1)!%P=P-1.所以答案就是(P-1)((P-1)...*(Q-1)的逆元)%P.数据很大,用__int128. 代码如下 #include <bits/stdc++.h> #define INF 0x3f3f3f3f using namespace std; typedef long long ll; inline int read(){ in

参考博客:https://blog.csdn.net/birdmanqin/article/details/97750844 题目链接:链接:http://acm.hdu.edu.cn/showproblem.php?pid=6608   威尔逊定理:在初等数论中,威尔逊定理给出了判定一个自然数是否为素数的充分必要条件.即:当且仅当p为素数时:( p -1 )! ≡ -1 ( mod p ),但是由于阶乘是呈爆炸增长的,其结论对于实际操作意义不大. 题意:T组样例.每组样例,给出一个素数P(1e

Fansblog Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 3170 Accepted Submission(s): 671 Problem Description Farmer John keeps a website called 'FansBlog' .Everyday , there are many people visite

Zball in Tina Town Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others) Problem Description Tina Town is a friendly place. People there care about each other. Tina has a ball called zball. Zball is magic. It grows lar

威尔逊原理.即对于素数p,有(p-1)!=-1( mod p). 首先,将原式变形为[ (3×k+6)! % (3×k+7) + 1] / (3×k+7),所以: 1.3×k+7是素数,结果为1, 2.3×k+7不是素数,则假设(3×k+7)=m1*m2*m3……,可知m1,m2,m3……<=3*k+6,则此时(3×k+6)! % (3×k+7) = 0,所以经过取整,式子的答案为0. #include<cstdio> using namespace std; ],sum[]; void

YAPTCHA Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2041 Accepted Submission(s): 1047 Problem Description The math department has been having problems lately. Due to immense amount of unsolic