reciprocity theorem

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=5446The main topic: C (n, m)% m, where M is the product of different prime numbers, namely M=P1*P2*...*PK, 1≤k≤10.1≤ m≤ n≤10^18. Analysis: If M is a prime number, it can be done directly with the Lucas theorem, but M is not a prime, but a multiplication of prime numbers. so that C (n, m) is x, you can use the Lucas theorem to calculate the X%p1,x%p2, ...

Title Link: BZOJ-3129Problem analysisThe idea of using the partition method, if there are no restrictions, then the program number is C (m-1, n-1).If there is a limitation of Xi >= Ai, then we can subtract the M from Ai-1, which is equivalent to a fixed portion of this part to Xi, which translates into unrestricted conditions.What if there are some restrictions on the XI Consider the allowance: consider which restrictions are violated, that is, what restrictions are XI Then we can find out the a

Topic Links:http://poj.org/problem?id=1286Main topic:Given 3 colors of beads, the number of each color beads are unlimited, the beads are made of the length of the necklace of N.Ask how many kinds of non-repeating necklaces can be made, and the final result will not exceed the range of int type data. and twoThe necklace is the same, when and only if the two necklaces by rotating or flipping can be coincident together, and the corresponding beadSame color.Problem Solving Ideas:The problem is the

　　we all know that. Euclidean algorithm is used to quickly find two number of greatest common divisor algorithm, high efficiency: 2O (LOGN). We first give the implementation of the algorithm:1 intGcd_1 (intAintb)2 {3 if(b==0)returnA;4 returnGcd_1 (b, a%b);5 }6 7 intGcd_2 (intAintb)8 {9 while(b)Ten { One intr = a%b; AA =b; -b =R; - } the returnA; -}View Code　　To prove the correctness of the above algorithm, we are proving the correctness of the following

Chinese remainder theorem againTime limit:1000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): Accepted submission (s): 777Problem description I know some of the students are looking at the Chinese remainder theorem recently, this theorem itself, or relatively simple:Assuming that the m1,m2,..., Mk 22, the following equation

Prime number theorem: A number of prime numbers that are recorded as less than or equal, then there aretheorem: set, and then there istheorem: set, and thentheorem: set, then the value is(1) for prime numbers, then the answer is(2) There are a number of factors, then the answer is(3) There is only one vegetarian factor, so the answer is that the vegetarian factortheorem: set to Fib number, then there istheorem: Given the positive integers of two, the

Triple peak occurs in 1234 days. Use the plural form ''days ''even if the answer is 1. Sample Input 0 0 0 00 0 0 1005 20 34 3254 5 6 7283 23 102 320203 301 40-1-1-1-1 Sample output Case 1: The next Triple peak occurs in 21252 days. case 2: The next Triple peak occurs in 21152 days. case 3: The next Triple peak occurs in 19575 days. case 4: The next Triple peak occurs in 16994 days. case 5: The next Triple peak occurs in 8910 days. case 6: The next Triple peak occurs in 10789 days.Chine

controlled by $ \ DPS {P (z) =\ VLS {n} C_n} $. according to chapter 7 of theorem 7, $ L $ can be controlled to extend to the whole $ B $. note \ eqref {4_2_eq}. We know $ \ Bex L (x) = \ lim _ {n \ To \ infty} a_n \ quad \ sex {x = (A_1, a_2, \ cdots) \ In B }. \ EEx $ 3. proof: there is a generalized limit of $ t \ To \ infty $, so that the definition of $ \ sed {T \ In \ BBR; \ t \ geq 0} $ all Bounded Functions on $ x (t) $, the generalized limi

Problem dMorley's TheoremInput:Standard Input Output:Standard output Morley's theorem states that the lines trisecting the angles of an arbitrary plane triangle meet at the vertices of an equilateral triangle. for example in the figure below the tri-sectors of angles A, B and C has intersected and created an equilateral triangle def. Of course the theorem has various generalizations, in the particle if al

Today, the subject of the exam has a large number of models, not alas, lost 45 points, I am really a weak chicken, and now will not Lucas.So I looked at it today, and the theorem is almost:(1) Lucas theorem: P is a prime number, then there are:That is:lUcas (n,m,p) =c (n%p,m%p) *lucas (n/p,m/p,p) 　　 and then left my understanding:To prove the formula in a recursive way;Consider factorial first, in the sense

Test instructions: On a plane, given a closed stroke graph consisting of n points (4 Euler's theorem needs to be used;Euler's theorem: the number of vertices in the graph is V, the number of sides (the number of edges in the three dimensions) is E, and the number of polygons is F, then v + f-e = 2.If the number of polygons is asked, only the number of vertices V and the number of sides e, that is, can be ob

Pick theorem, pick theorem is like this, S = a + B/2-1, where S is the graph area, a is the number of cells in the graph, and B is the number of cells passing through the edge, applicable scope: vertices are all integral points, or simple polygon of vertices. How to find the area? The multiplication of triangles. This is depressing because the vertex in the figure is not the vertex obtained by using the pi

Codeforces 464C Substitutes in Number homophone theorem + simulation, codeforces464c Question link: Click the open link Question: Given a string of numbers There are n operations below The format of each line is like d-> t. D is a digit and t is a number of any length. T Length and length cannot exceed 100000 Q: The final result is % 1e9 + 7. Ideas: First, we can draw a conclusion: The number cannot be modified after the same

The Chinese proofs of the theoremble theorem circulated on the Internet are not so easy to understand. So here I will describe my general proof. Schmerlot's theorem: in a clear-cut chess game (such as Chinese chess and chess) that will surely end in a limited number of steps. In any particular situation, set a first-to-second, after B, either a will win the next method, B will win the next method, or A and

contaminated, each method gives an AI, a pi, if the lucky number%pi equals AI will be contaminated, and then give you an interval,Ask how many lucky numbers have not been contaminated in this interval. The problem: This question was first known to be a Chinese residual theorem, and then read a number of problems, found to use the inclusion of the Chinese remainder theorem, and then found to explode long lo

theorem Looking back on the Burnside algorithm above, does it feel that D is not very well implemented?Then we can use a method that is easier to understand, adaptable, time complexity low, programming complexity is the most powerful polya theorem of group theoryIn fact, the difference between Polya theorem and Burnside lemma is not very big, but the angle of th

We define Ψ (x) as Euler functions. Euler theorem Description: If (A, p) = 1 , then aψ (P) ξ1 (mod p) . Prove: Warm up first:Ⅰ. We make x1, x2, x3, ..., xs a simplified residual system for modulo p (if for any 1≤j≤s, (XJ, p) = 1 and for any a∈z, if (A, p) = 1 , then there is only one XJ that is the remainder of a to modulo p (and xI 22 is not the same).) where s =ψ (p) .Ⅱ. If {x1,x2,x3,......, xs} is a simplified rem

Analysis: The derivation of the Lucas Theorem requires C (n,m)%2, then the Lucas theorem, written as binary observation, such as N=1001101,m is an enumeration from 000000 to 1001101, in the theorem C (0,1) = 0, so n= 1001101 of the 0 corresponds to the position of M bits is 1 then C (n,m)% 2==0, so m corresponding to the position of N 0 can only fill 0, and 1 of

Ask G's P-%mod,According to the Fermat theorem, G^sigma (c (n,d)) (d|n)%mod=g^ (Sigma (C (n,d)) (d|n)% (mod-1))%mod,However, Mod-1 is not a prime number, it can only be used to split it into 4 mass factors, and then to solve the 4 equations respectively, first with the Lucas theorem and Ma Xiaoding to find out the value of 4 prime numbers of Sigma (Num[i]), notice that the enumeration factor d is enumerated

1997 Wu Wenjun Academician won the Herbrands award, what is the explanation? First of all, we must be clear: The Herbrands Award is the highest academic award in the field of automatic theorem proving in the international mathematics world today. The award marks the unanimous endorsement of the award-winning project by the International academic community (Tongren). Looking back over the years, the list of Herbrands winners is as follows:Larry Wos (1