LA 4119 (differential sequence polynomial) always an integer

Test instructions

A polynomial with a shape such as (p)/d is given, where P is the integer coefficient polynomial of N and D is an integer.

Ask if the value of the polynomial is an integer for all positive integer n.

Analysis:

It can be proved by mathematical induction that if P (n) is a k-th polynomial, p (n+1)-P (n) is a K-1 quadratic polynomial.

P is a polynomial of N, p is a arithmetic progression, as long as the validation p (1) and P (2) are multiples of d.

P is a two-time polynomial of N, as long as the first item is a multiple of D, and the difference between two adjacent items is a multiple of d. The difference between the two adjacent items is a polynomial, so to verify the two items, together with the first item of the previous validation, the total validation of P (1), P (2), and P (3) three items.

In general, to verify the K-th polynomial, just verify that P (1) ... P (k+1).

The value of the polynomial can be calculated using the Qin Jiushao algorithm that the high school math textbook has talked about.

1#include <iostream>2#include <cstdio>3#include <string>4#include <cstring>5#include <cstdlib>6 using namespacestd;7 8 Const intMAXN = -+Ten;9 intA[MAXN], p;//A[i] represents the n^i corresponding coefficient, p is the highest coefficientTen  One voidParse_polynomial (strings) A { -     inti =0, Len =s.size (); -      while(I <len) the     { -         intSign =1; -         if(S[i] = ='+') i++; -         if(S[i] = ='-') {i++; sign =-1; } +         intv =0;//coefficient -          while(I < len && isdigit (S[i])) v = v *Ten+ s[i++]-'0'; +V *=Sign ; A         if(i = len) a[0] = V;//Constant Entry at         Else -         { -             if(v = =0) v =Sign ; -             intU =1;//No index -             if(S[++i] = ='^')//have index -             { inU =0; -i++; to                  while(IsDigit (s[i])) u = u *Ten+ s[i++]-'0'; +             } -A[u] =v; the             if(U > P) p =u; *         } $     }Panax Notoginseng } -  the intMoDintXintMOD) + { A     intAns =0; the      for(inti = P; I >=0; i--) +     { -Ans = (Long Long) ans * x% MOD;//Be careful not to overflow $Ans = ((Long Long) ans + a[i])%MOD; $     } -     returnans; - } the  - BOOLCheckstring&expr)Wuyi { the     intp = Expr.find ('/'); -Parse_polynomial (Expr.substr (1, P-2)); Wu     intD = Atoi (Expr.substr (p+1). C_STR ()); -      for(inti =1; I <= p+1; i++) About         if(MoD (i, D)! =0)return false; $     return true; - } -  - intMain () A { +     //freopen ("In.txt", "R", stdin); the  -     intKase =1; $     stringexpr; the      while(Cin >>expr) the     { the         if(expr[0] =='.') Break; theMemset (A,0,sizeof(a)); -p =0; inprintf"Case %d:%s\n", kase++, check (expr)?"Always an integer":"Not always an integer"); the     } the  About     return 0; the}

code June

LA 4119 (differential sequence polynomial) always an integer