polynomial calculator

Polynomial multiplication optimization algorithm:The following two polynomial are available:Make a vector of their coefficients x=[x0,x1,x2,x3,......] The form, to bef=[2,3,1] g=[5,2,0]So according to the convolution formulaConvolution of vectors f and g can be obtained s=[10,19,11,2]The polynomial multiplication can be used to calculateThe coefficients and the a

1010. One-element polynomial derivation (25) time limit of MS Memory Limit 65536 KB Code length limit 8000 B The Standard of the Judgment procedure The design function asks for the derivative of the unary polynomial. (Note: The first derivative of xn (n is an integer) is n*xn-1. ） Input format: Enter the polynomial non 0 coefficients and indices (integers with an

Reprint Please specify source: http://blog.csdn.net/lsh_2013/article/details/46697625Least squares (also known as the least squares method) is a mathematical optimization technique. It matches by minimizing the squared error and finding the best function of the data.The C + + implementation code is as follows:#include Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced. The principle of least squares fitting

C language: calculate the value of Polynomial 1-1/2 + 1/3-1/4 +... + 1/99-1/100, three types of cyclic implementation Method 1: for Loop Implementation Program: # include

Test instructions: Given a string, there are two operations, one is p a B, the string from position A to the substring of position B is a palindrome substring, the other operation c a B, the character of the string position A is replaced by B.Solution: Because the string length is 1e5 and the number of times to ask also has 1e5, so the violence must be timed out, and then consider using a tree array to maintain the hash value of the string to solve, two operations with positive and negative dire

] = $derivative;$max _derivative = $max _derivative>=abs ($derivative)? $max _derivative:abs ($derivative);printf ("x=%f, derivative=%f \ n", $x _data, $derivative);}$matchs = Array ();foreach ($derivatives as $x _data=> $derivative) {if (ABS ($derivative) = = $max _derivative) {$matchs [] = $x _data;}}printf ("Max derivative=%f\n", $max _derivative);foreach ($matchs as $x _match) {printf ("Derivative=%f when x=%f\n", $derivatives [$x _match], $x _match);}}Notice the format of formula:ax^b if b=

PHP write the polynomial derivative of the function code, the need for friends can refer to the next The output would be:Sample x values:-12,-11,-10,-9,-8,-7,-6,-5,-4,-3,-2,-1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12,Max derivative=26.000000derivative=26.000000 when x=12.000000 The above is the whole content of this article, I hope that everyone's learning has helped, more relevant content please pay attention to topic.alibabacloud.com!

; ':Hb->next = hb->next->next; Delete the node the QB points toQb->next = ha->next; Before inserting the QB into HA QAHa->next = QB; QB = hb->next; Not QB = Ha->nextHa = ha->next;Break Defaultcout Break}}if (QB)Ha->next = QB;Free (HB);} void Main (void){Node *a = NULL;Node *b = NULL;int CountA;int COUNTB;cout Creatpoly (A, CountA); Generate a linked listcout Cin >> COUNTB;Creatpoly (B, COUNTB); cout Showpoly (A);cout Showpoly (B); Addpolynomial (A, B); A = a + Bcout Showpoly (A); The sum of the

;exp); p = p->link; }}ostreamoperator out,ConstPolynominal x) {X.output ( out); return out;} IStreamoperator>> (IStream inch, Polynominal x) {X.addterms (inch); return inch;} Polynominaloperator+ (Polynominal a, Polynominal C) {A.polyadd (b); returnA;}voidPolynominal::P olymult (Polynominal R) { term* Q, *q1=thelist, *p,*mult; P= r.thelist->link; Q= q1->link; Polynominal temp; intc =0, E =0; Mult= temp.thelist->link; for(q = r.thelist->link; p->exp >=0; p = p->link) { for(; Q->ex

Test instructions(x1 + x2 + x3 +...+xk) ∧n The coefficients of a particular item ...Code#include typedefLong Longll;ll C (intNintk) {ll ans=1; inttemp =0; while(temp! = k) {ans *= n-temp; temp++; } for(intI=1; ii; returnans;}intMain () {intex, M; while(SCANF ("%d%d", ex, m) = =2) { intarray[ the]; for(intI=0; i"%d", Array[i]); ll Res=1;intnow =ex; for(intI=0; iif(Array[i]) {res*=C (now, array[i]); now-=Array[i]; } printf ("%lld\n", RES); } return 0;}uva10105-

$v$. Proof: Exists $\alpha_{1},\alpha_{2},..., \alpha_{n}\in \mathbb{f}$ makes $$ \sum_{k=1}^{n}\alpha_{k} (T-\lambda_{k}i) ^{-1}=I. $$ Proof: 1. For any $i=1,2,..., n$, $T-\lambda_{i}$ irreversible $\iff \left| T-\lambda_{i}\right|=0\iff \lambda_{i}$ is the characteristic value of $t$, and by the condition of the question set, $\lambda_{i}$ is not the characteristic value of $t$, so $t-\lambda_{i}$ is reversible.2. Make $ $F (x) =\prod_{k=1}^{n} (X-\lambda_{k}) $$ because $t-\lambda_{

#include Data structure learning note solving polynomial 2

#include #includestring.h>structn{intC; inte;} buf[1010],ans[1010];intMain () {Freopen ("data.in","R", stdin); Freopen ("D1.out","W", stdout); intx,y,index=0, k=0, I; while(SCANF ("%d%d", x,y)! =EOF) {BUF[INDEX].C=x; BUF[INDEX].E=y; if(buf[index].e==0) Break; Index++; }k=index;if(k==0) printf ("0 0");Else { for(i=0; i1; i++) { if(buf[i].e!=0) printf ("%d%d", buf[i].c*buf[i].e,buf[i].e-1); } printf ("%d%d", buf[i].c*buf[i].e,buf[i].e-1); } return 0;}Derivation of one-element

The coefficient of x1n1x2n2... xknk is (n, N1) (n-n1, N2)... (n-n1-n2-...-NK-1, NK) = n! /N1! N2 !... NK !. Code: /*************************************** ********************************** * Copyright (c) 2008 by liukaipeng * * Liukaipeng at gmail dot com * **************************************** *********************************/ /* @ Judge_id 00000 10105 C ++ "polynomial coefficients "*/ # Include # Include # Include # Include

Polynomial multiplication1#include 2 using namespacestd;3 Const DoublePI = ACOs (-1);4 Const intLEN = 1e5 +5;5 intN, M, N;6 namespacePloy {7 structComp {8 Doublex, y;9} A[len *4], B[len *4];TenCompoperator+ (ConstComp x,ConstComp y) { One return(comp) {x.x + y.x, x.y +y.y}; A } -Compoperator- (ConstComp x,ConstComp y) { - return(comp) {x.x-y.x, x.y-y.y}; the } -Compoperator* (ConstComp x,ConstComp y) { - ret

);qb=hb->next;ha=ha->next;Break}} if (!pb->next) append (PA,QB); Free (HB);}int main (void) {int NUMBEROFPA,NUMBEROFPB;Polyn *pa=null,*pb=null,*p=null;scanf ("%d%d", NUMBEROFPA,NUMBEROFPB);Pa=creatpolyn (PA,NUMBEROFPA);printf ("List A is complete \ n");p=pa->next;while (P!=null) {printf ("%.2f%d\n", P->COEF,P->EXPN);p=p->next;}Pb=creatpolyn (PB,NUMBEROFPB);printf ("List B is set \ n");Addpolyn (PA,PB);printf ("addition complete \ n");while ((Pa->next)!=null) {pa=pa->next;printf ("%.2f%d\t", PA->

Problem DescriptionEnter a 1 positive integer n,Calculation 1+ (1+2) + (1+2+3) +...+ (1+2+3+...+n)InputInput positive integer n (multiple sets of data)OutputOutput 1+ (1+2) + (1+2+3) +...+ (1+2+3+...+n) value (one row per group of data)Sample Input2Sample Output41#include 2 3 using namespacestd;4 5 intMain ()6 7 {8 9 intn,t;Ten One Longsum; A - while(cin>>N) - the { - -sum=0; - +t=0; - + for(intI=1; i) A at { - -

] +β2*[radio]F (Sales) = 2.9211 + 0.0458 * [TV] + 0.188 * [radio] In the figure, in the sales direction, the blue point is the original sales actual value, and the red point is the value calculated by the Fit function. In fact, the error is not big, some of the data below.Also can be quasi-unity meta-function;ImportNumPy as NPImportPandas as PDImportStatsmodels.formula.api as SMFImportMatplotlib.pyplot as Plt fromMpl_toolkits.mplot3dImportAXES3DDF= Pd.read_csv ('Http://www-bcf.usc.edu/~gareth/I

The example of this article is about the simple realization of C language to find the N-order polynomial method. Share to everyone for your reference, specific as follows: #include I hope this article will help you with the C language program design.

Description:M m amino acids, known for their relative molecular mass of c1,c2,c3 ... C 1, C 2, c 3 ... C1,c2,c3., after a precise dehydration condensation, a large number of various peptide chains are formed. It is necessary to predict how many polypeptide chains have been hydrolyzed to the relative molecular mass and to n n N. (A−b−c (A−b−c (a-b-c and c−b−a c−b−a c-b-a two peptide chains as different) Solution:Consider having all the elements built into a generating function a (x) a (x) a (x),