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>…

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…

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)…

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…

---恢复内容开始--- 10．3二重积分的换元积分法 在一元函数定积分的计算中,我们常常进行换元,以达删繁就简的目的,当然,二重积分也有换元积分的问题. 首先让我们回顾一下前面曾讨论的一个事实. 设换元函数 ,视其为一个由定义域到的映射．点的象点为,点x的象点为,记 , 则由到点的线段长为,到的线段长为,称为映射在点到点的平均伸缩率.若在点处可导,则 = 即称是映射在点处的伸缩率. 对于由平面区域到的映射我们有如下结论: 引理 若变换在开区域存在连续偏导数,且雅可比行列式,.变换将平面上开区域…

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…

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…

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…

<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…