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…

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…

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…

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…

f(n) Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 457    Accepted Submission(s): 279 Problem Description This time I need you to calculate the f(n) . (3<=n<=1000000) f(n)= Gcd(3)+Gcd(4)+…+Gc…

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…

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…

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…

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,…

Kingdom of Obsession Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 200    Accepted Submission(s): 64 Problem Description There is a kindom of obsession, so people in this kingdom do things ver…

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…

<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)…

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…

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…

题意: 输入n,求c(n,0)到c(n,n)的所有组合数的最小公倍数. 输入: 首行输入整数t,表示共有t组测试样例. 每组测试样例包含一个正整数n(1<=n<=1e6). 输出: 输出结果(mod 1e9+7). 感觉蛮变态的,从比赛开始我就是写的这道题,比赛结束还是没写出来…… 期间找到了逆元,最小公倍数,组合数的各种公式,但是爆了一下午tle. 比赛结束,题解告诉我,公式秒杀法…… 但是公式看不懂,幸好有群巨解说,所以有些听懂了,但还是需要继续思考才能弄懂. 题解: 设ans[i]表示i…

Invoker Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 122768/62768K (Java/Other) Total Submission(s) : 1   Accepted Submission(s) : 0 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description On of Vance's favourite hero i…

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…

#include <cstdio> #include <ctime> #include <cmath> #include <algorithm> using namespace std; typedef long long ll; const int N = 108; const int S = 10; ll mult_mod(ll a, ll b, ll c) { a %= c; b %= c; ll ret = 0; while(b) { if(b&am…

题目 "在那山的那边海的那边有一群小肥猪.他们活泼又聪明,他们调皮又灵敏.他们自由自在生活在那绿色的大草坪,他们善良勇敢相互都关心--" --选自猪王国民歌 很久很久以前,在山的那边海的那边的某片风水宝地曾经存在过一个猪王国.猪王国地理位置偏僻,实施的是适应当时社会的自给自足的庄园经济,很少与外界联系,商贸活动就更少了.因此也很少有其他动物知道这样一个王国. 猪王国虽然不大,但是土地肥沃,屋舍俨然.如果一定要拿什么与之相比的话,那就只能是东晋陶渊明笔下的大家想象中的桃花源了.猪王勤政爱…

首先我按着我的理解说一下它为什么是卡特兰数,首先卡特兰数有一个很典型的应用就是求1~N个自然数出栈情况的种类数.而这里正好就对应了这种情况.我们要满足题目中给的条件,数字应该是从小到大放置的,1肯定在左上角,所以1入栈,这时候我们放2,如果我们把2放在了1的下面就代表了1出栈,把2放在上面就代表了2也进栈(可以看一下hint中第二组样例提示),以此类推,这样去放数,正好就对应了上面一行入栈,下面一行出栈的情况,一共n行,对应上限为n的卡特兰数. 需要注意的地方就是在使用卡特兰数递推式的时候,除法…

HDU 2012 素数判定 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 72727    Accepted Submission(s): 25323 Problem Description 对于表达式n^2+n+41,当n在(x,y)范围内取整数值时(包括x,y)(-39<=x<y<=50),判定该表达式的值是否都为素数.…

Saving Beans Saving Beans Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5761    Accepted Submission(s): 2310 Problem Description Although winter is far away, squirrels have to work day and nig…

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…

NPY and girls Problem Description NPY's girlfriend blew him out!His honey doesn't love him any more!However, he has so many girlfriend candidates.Because there are too many girls and for the convenience of management, NPY numbered the girls from 1 to…

题目链接:Lucky7 题意:求在l和r范围内,满足能被7整除,而且不满足任意一组,x mod p[i] = a[i]的数的个数. 思路:容斥定理+中国剩余定理+快速乘法. (奇+ 偶-) #include <stdio.h> #include <string.h> #include <iostream> using namespace std; #define LL long long #define FOR(i, n) for (int i=0; i<n; +…