stirling engine

First class Stirling number S (p,k) A combinatorial interpretation of S (p,k) is the number of methods that arrange P objects into K-non-empty loops. Recursive formula for S (p,k): s (p,k) = (p-1) *s (p-1,k) +s (p-1,k-1), 1 Boundary conditions:s (p,0) =0, p>=1 S (p,p) =1, p>=0 Description of recurrence Relationship: Considering the P item, p can constitute a non-empty loop arrangement, so that the former p-1 items constitute k-1 loops, and the me

Note that this is about the Stirling number, not the Sterling formula.There are two types of sterling: the first class of Stirling and the second class of Stirling. be recorded separately.First, the second class of Stirling numbers is described.Description: The number of methods that divide a collection of n items into

Tags: acm algorithm combination mathematical stiring countI. The second type of Stirling numberTheorem: The second type of Stirling number S (p, k) counts the number of Division of the p element set into k undistinguished boxes without empty boxes.Proof: it is not important for elements to take some boxes. The only thing that matters is what is installed in each box, no matter which box is loaded.Recursive

HDU 2643/*The second type of Stirling is the number of methods that divide a set containing n elements into k non-empty subsets.Recurrence formula:S (n, k) = 0 (n S (n, n) = S (n, 1) = 1,S (n, k) = S (n-1, k-1) + KS (n-1, K ).*/ # Include HDU 2512 # Include TIPS: Bell number, also known as Bell number.Eric temple bell is the name of Eric temple bell. B (n) is the number of division methods for a set containing n elements. B (0) = 1, B (

Bell (hdu4767 + matrix + Chinese Remainder Theorem + bell number + Stirling Number + Euclidean), hdu4767stirlingBellTime Limit:3000 MSMemory Limit:32768KB64bit IO Format:% I64d % I64uSubmit Status Practice HDU 4767 DescriptionWhat? MMM is learning Combinatorics !?Looks like she is playing with the bell sequence now:Bell [n] = number of ways to partition of the set {1, 2, 3,..., n}E.g. n = 3:{1, 2, 3}{1} {2 3}{1, 2} {3}{1, 3} {2}{1} {2} {3}So, bell [3

Introduction and distinction of two types of Stirling number (refer to csdn from Acdreamer)Stirling number I ---s (n,k): A method that ranks n objects into a K non-empty loop arrangement (ring).Recursive: S (n, k) = (n-1) *s (n-1, K) + S (n-1, k-1); 1Explanation: Consider the n+1 element 1, the formation of a circular arrangement alone, the remainder of the S (n-1, k-1) species method2, and other elements t

The Stirling formula is a mathematical formula used to obtain an approximate value of n factorial. Generally speaking, when N is very large, n factorial calculation is very large, so the Stirling formula is very useful, and even when N is very small, the value of the Stirling formula is very accurate. The formula is: The programming of the

The second type of Stirling is the number of methods that divide a set containing n elements into k non-empty subsets.Recurrence formula:S (n, k) = 0 (n S (n, n) = S (n, 1) = 1,S (n, k) = S (n-1, k-1) + kS (n-1, k ). The following figure is taken from Wikipedia. There is a conversion formula for the binary Stringer number and the combination number. However, it is easy to combine the remainder of the number 2. We know that the combination number C (N

Topic Links:http://acm.hdu.edu.cn/showproblem.php?pid=2512Main topic:There are n cards, the n card is divided into several different sets, the collection cannot be empty. Q: How many kinds of sub-methods are there altogether?Ideas:Reference post: http://blog.csdn.net/acm_cxlove/article/details/7857671The number of collections can be 1, 2, 3 、...、 N. The problem changed to put n cards in the I set.This time a combinatorial problem can be solved with the second type of

Fifth session of the provincial race: HearthstoneCombinatorial mathematics.n Races, M tables (n>=m). Each race is a table, and each table is used at least once.The idea behind the question has been how to fill the M table with n positions.is actually the Tao Stirling number model, the direct set formula M!*{n m}#include #includestring.h>#defineL 1000000007intMain () {intn,m; Long Longa[101]; while(SCANF ("%d%d", n,m)! =EOF) {a[0]=0; for(intI=1; i) A

as spaces, very space, they are inserted into this r element of the R+1 neutral, This will ensure that the enumeration of the number of methods known, cut to meet the random two element number difference >=k requirements So the problem is reduced to: assigning space elements to r+1 regions, which can be empty The answer is: stir2[space][r+1], the second kind of Stirling number (2) Then, the selected R elements are divided into no more than

The first class of Stirling number is positive or negative, and its absolute value is the number of methods that contain n elements that are arranged as K-rings.The recursive formula is, S (n,0) = 0, S (= 1). S (n+1,k) = S (n,k-1) + NS (n,k).Boundary conditions: s (0, 0) = 1 s (p, 0) = 0 p>=1 s (p, p) =1 p>=0Some properties: S (P, 1) = 1 p>=1 s (P, 2) = 2^ (p-1) –1 p>=2The second class of Stirling is the nu

Title Link: http://lightoj.com/volume_showproblem.php?problem=1326Baidu Encyclopedia: Sterling NumberAcdreamer: The first class of Stirling and the second class StirlingDisky and Sooma, and both of the biggest mega minds of Bangladesh went to a far country. They ate, coded and wandered around, even in their holidays. They passed several months in the this. But everything have an end. A Holy person, Munsiji came to their life. Munsiji took them to Derb

partitions that contain a collection of n+1 elements, considering the elementsAssuming that he is divided into one category alone, then there are still n elements, in which case the number is divided;Assuming that he and an element are divided into a class, then there are n-1 elements left, in which case the number is;Assuming that he and some two elements are divided into a class, then there are still n-2 elements, in this case the number is divided;And so on, we get the combination formulaThe

Problem description A set of n elements {1, 2,..., n} can be divided into several non-empty subsets. For example, when n = 4, the set {1, 2, 3, 4} can be divided into 15 different non-empty subsets as follows: {1}, {2}, {3}, {4 }}, {1, 2}, {3}, {4 }}, {1, 3 }, {2}, {4 }},{ {}, {2}, {3 }},{ 2, 3}, {1}, {4 }}, {2, 4}, {1}, {3 }}, {3, 4}, {1}, {2 }}, {1, 2}, {3, 4 }}, {1, 3}, {2, 4}, {1, 4}, {2, 3}, {4 },{ 1, 2, 4 }, {3 }},{ 1, 3}, {2 }},{ 2, 3, 4}, {1 }},{ 1, 2, 3, 4 }} Given a positive integer

Recently a pile of questions to fill, has been salted fish, fill a pile of water problems are not necessary to write a problem. Forget about this formula.The meaning of the Stirling formula is that when n is large enough, n! is very difficult to calculate, although there are many equations about n!, but it is not very good to estimate factorial results, especially after N is large, the error will be very large. However, the

] to indicate number I is the number of programs of JDp[i][j]= (Dp[i][j]+dp[z][j-1])%mod (z = number from 1 to i-k)But this requires a triple loop that takes more time than the data rangeAnd this is actually only related to the former (i-k), so we need to preprocess it in a rolling array.Step two: If you just put the M box, it's the second class of Stirling. But here is not much more than M boxes, so as a second class of

With EOF.Input would be a four integers n,r,k,m.we assume that they is all between 1 and 1000.Outputoutput the Maxmium days modulo 1000000007.Sample INPUT5 2 3 2Sample OUTPUT6HintSample input means you can choose 1 and 4,1 and 5,2 and 5 in the same day. And you can make the machines in the same group or the different group. Got 6 schemes.1 and 4 in same group,1 and 4 in different groups.1 and 5 in same group,1 and 5 in different groups.2 and 5 in same group,2 and 5 in different groups. We assum

, very space, they are inserted into the r+1 of the R element in a neutral, This guarantees the enumeration of the number of methods that are known, and the requirement to meet any two element number difference >=k So the problem is reduced to: assigning space elements to r+1 zones, which can be empty The answer is: stir2[space][r+1], the second kind of Stirling number (2) Then, the selected R elements are divided into no more than G Group,

Topic Link: Click to open the linkSterling Number: Click to open linkTest instructions is calculated n! The number of digitsThat is, ans = log10 (n!) = log10 (sqrt (2πn)) + N*LOG10 (n/e)#include HDU 1018 Big number Stirling approximate n!