Simple Number Theory

Time Limit: 3000/1000 ms (Java/other) memory limit: 65535/32768 K (Java/other) total submission (s): 15 accepted submission (s): 5 Font: times New Roman | verdana | georgiafont size: Drawing → problem descriptiongiven two positive integers A and N, satisfaing gcd (A, n) = 1, please find the smallest positive integer x with a ^ x limit 1 (mod N ). inputfirst is an integer T, indicating the number of test cases. T <3001.
Each of following T lines contains two positive integer a and N, separated by a space. A <n <= 1000000. outputfor each test case print one line containing the value of X. sample input

22 33 5

Sample output

24

Authorhynu // question: Find the minimum x value that satisfies a ^ x mod n constant = 1. // Background: For any two positive integers a, m, m> = 2, there is a ^ PHI (m) 1_1 (mod m) PHI (m) that is, the Euler's function of M, that is, the Euler's theorem. When M is a prime number P, this formula is: a ^ (p-1) 1_1 (mod m) it indicates that if M is a prime number, then the Euler's function of M is expressed as m-1. That is, ferma's theorem. // Question: First name the values of the 1-N Euler's function and then enumerate the Euler's function for prime factor decomposition to continuously update the minimum value.

# Include <iostream> # include <cstdio> # include <cmath> # include <cstring> using namespace STD; const int max = 1000010; int prime [Max], phi [Max]; // Euler's function for saving all values void fun () // evaluate the Euler's function for all values from 1 to max {Prime [0] = prime [1] = 0; For (INT I = 2; I <= max; I ++) prime [I] = 1; for (INT I = 2; I * I <= max; I ++) if (prime [I]) for (Int J = I * I; j <= max; j + = I) prime [J] = 0; For (INT I = 1; I <= max; I ++) Phi [I] = I; for (INT I = 2; I <= max; I ++) if (prime [I]) f Or (Int J = I; j <= max; j + = I) Phi [J] = Phi [J]/I * (I-1);} int Mod (int, int B, int c) // fast idempotent modulo {int ans = 1; long AA = A; while (B) {If (B % 2) ans = ans * Aa % C; AA = AA * Aa % C; B/= 2;} return ans;} int main () {// freopen ("a.txt ", "r", stdin); // freopen ("B .txt", "W", stdout); fun (); int T, A, N; CIN> T; while (t --) {CIN> A> N; int Sn = (INT) SQRT (PHI [N]), ANS = N; // enumerate the Euler's factor of N between 1-sn For (INT I = 1; I <= Sn; I ++) // find all the factors in the Euler's function that meet the conditions and are the smallest. {If (PHI [N] % I = 0) // If I is a factor of Phi, update the minimum value {If (mod (A, I, n) = 1 & Ans> I) ans = I; If (mod (A, Phi [N]/I, n) = 1 & Ans> Phi [N]/I) ans = Phi [N]/I ;}} cout <ans <Endl ;}return 0 ;}

 

Simple Number Theory