51nod 1135 Original Root

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Set M is a positive integer, A is an integer, if the order of a-mode m is equal to φ (m), then A is called a primary root of modulo m. (where φ (m) represents the Euler function of M)

Order: gcd (a,m) = 1, which makes the smallest restablished, called A-to-modulo m-order .

φ (m): The number of coprime in the interval between [1,m] and M.

The method for finding the original root A of modulus p:

because P is a prime number, so φ (p) =p-1, the problem is to find the smallest a so that a^ (p-1)%p = 1 established (according to the Fermat theorem, the formula must be established),

First of all, p-1 all the different quality factor p1,p2...pm,

For any integer a∈[1,p-1], verify whether a is the root of P,

Test method: a^ ((p-1)/p1), a^ ((p-1)/p2),... a^ ((p-1)/pm) There is a modulo p equal to 1,

If a is present, a is not a root of modulo p (that is, p-1 is not a-to-modulo p-order), otherwise a is the original root.

1#include <cstdio>2#include <cstring>3#include <algorithm>4#include <vector>5#include <queue>6 #defineCLR (A, B) memset ((a), (b), sizeof ((a)))7 using namespacestd;8typedefLong Longll;9 Const intN =50000;Ten intPrime[n];//Prime[0] The number of primes that are stored One intPpri[n];//the quality factor of p-1 A voidGetprime () { -The CLR (Prime,0); -      for(inti =2; i < N; i++){ the         if(!Prime[i]) -prime[++prime[0] ] =i; -          for(intj =1; J <= prime[0] && Prime[j] <= n/i; J + +){ -prime[Prime[j] * I] =1; +             if(i% prime[j] = =0) Break; -         } +     } A } at ll Pow_m (ll A, ll N, ll m) { -ll t = a%m; -ll r =1; -      while(N >0){ -         if(N &1) -R = R * t%m; int = t%m; -N >>=1; to     } +     returnR; - } the intDivideintN) {//decomposition of the mass factor of composite n *     intCNT =0; $      for(inti =1; Prime[i] * Prime[i] <= N; ++i) {Panax Notoginseng         if(n% prime[i] = =0){ -PPRI[++CNT] =Prime[i]; the              while(n% prime[i] = =0){ +N/=Prime[i]; A             } the         } +     } -     if(N >1) ppri[++cnt] =N; $     returnCNT; $ } - intMain () { -     intp, I, A, T, F; the getprime (); -scanf"%d", &p);Wuyi     intCNT = Divide (P-1);//the number of qualitative factors in p-1 the      for(A =2; A <= P-1; ++a) {//original root from 2 to P-1 enumeration -f =1; Wu          for(i =1; I <= CNT; ++i) { -t = (P-1) /Ppri[i]; About             if(Pow_m (A, t, p) = =1){ $             //exists a^ ((p-1)/ppr[i]) mod p = 1, then A is not the original root of prime number p -f =0; Break; -             } -         } A         if(f) { +printf"%d\n", a); the              Break; -         } $     } the     return 0; the}

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51nod 1135 Original Root