UVa11287 - Pseudoprime Numbers(欧拉筛选法、快速求幂法)

 

Fermat's theorem states that for any prime number p and for any integer a > 1, a^p == a (mod p). That is, if we raise a to the pth power and divide by p, the remainder isa. Some (but not very many) non-prime values of p, known as base-a pseudoprimes, have this property for some a. (And some, known as Carmichael Numbers, are base-a pseudoprimes for all a.)

Given 2 < p ≤ 1,000,000,000 and 1 < a < p, determine whether or not p is a base-a pseudoprime.

Input contains several test cases followed by a line containing "0 0". Each test case consists of a line containing p and a. For each test case, output "yes" if p is a base-a pseudoprime; otherwise output "no".

Sample Input

3 210 3341 2341 31105 21105 30 0

Output for Sample Input

nonoyesnoyesyes

用欧拉筛选法求出[1,31623]之间的素数，对与p在这个范围内的，可以直接判断是不是素数，大于该范围的，用p除以[1,31623]之间的素数

用快速求幂算法求mod

import java.io.FileInputStream;import java.io.BufferedReader;import java.io.BufferedInputStream;import java.io.InputStreamReader;import java.io.StreamTokenizer;import java.io.PrintWriter;import java.io.OutputStreamWriter;import java.util.Scanner;import java.util.ArrayList;public class Main implements Runnable{private static final boolean DEBUG = false;private static final int N = (int)Math.sqrt(1000000000) + 1;private Scanner cin;private PrintWriter cout;private int p, a;private ArrayList<Integer> vPrime = new ArrayList<Integer>();private boolean[] vis;private void init() {try {if (DEBUG) {cin = new Scanner(new BufferedInputStream(new FileInputStream("e:\\uva_in.txt")));} else {cin = new Scanner(new BufferedInputStream(System.in));}cout = new PrintWriter(new OutputStreamWriter(System.out));sieve();} catch (Exception e) {e.printStackTrace();}}private void sieve(){vis = new boolean[N];vis[0] = vis[1] = true;for (int i = 2; i < N; i++) {if (!vis[i]) {vPrime.add(i);}for (int j = 0; j < vPrime.size() && i * vPrime.get(j) < N; j++) {vis[i * vPrime.get(j)] = true;if (i % vPrime.get(j) == 0) break;}}}private long exp(long a, long n, long b){long product = 1;while (n != 0) {if ((n & 1) != 0) product = product * a % b;a = (a % b) * (a % b) % b;n >>= 1;}return product;}private boolean input() {p = cin.nextInt();a = cin.nextInt();if (p == 0 && a == 0) return false;return true;}private boolean check(int n){if (n < N) return vis[n];for (int i = 0; i < vPrime.size(); i++) {int tmp = vPrime.get(i);if (n % tmp == 0) return true;}return false;}private void solve() {if (check(p) && exp(a, p, p) == (long)a) cout.println("yes");else cout.println("no");cout.flush();}public void run(){init();while (input()) {solve();}}public static void main(String[] args) {new Thread(new Main()).start();}}