http://acm.hdu.edu.cn/showproblem.php?pid=3307 (欧拉 + 推导公式 + gcd)

/* * 解题： 题目描述很简单的说，然后我们可以推得这样一个公式： an = x^n * a0 + y*sigma(x+x^2+....+x^(n-1) ) ---> an = x^n*a0 + y/(x-1)*(x^n-1);因为题目 给定 y mod (x-1) = 0 ;那么 如果an % a0 == 0 ; 那么 可以表示 ( y/(x-1)*(x^n-1) )%a0 == 0;设 m = y/(x-1), g = gcd(a0,m),那么我们可以得到 ( x^n-1)%(a0/g) == 0;设c = a0/g;转化为 x^n == 1(mod c );那么我们可以得到 当 gcd( c,x ) != 1 时 无解当gcd( c,x ) == 1 时，我们可以得到 x^eular(c) == 1 ( mod c);但是这不一定是最小的，所以我们要枚举x的约数，求得最小的。 */import java.io.*;import java.util.*;import java.math.*;public class Main3307{static final int N = 100000;static long num[] = new long[N];//public static void main(String[] args){public static void main(String[] args)throws IOException{StreamTokenizer cin = new StreamTokenizer(new BufferedReader (new InputStreamReader(System.in)));long x, y, a;PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));//Scanner cin = new Scanner(System.in);//while(cin.hasNext()){while(cin.nextToken() != StreamTokenizer.TT_EOF){//x = cin.nextLong();//y = cin.nextLong();//a = cin.nextLong();x = (long)cin.nval;cin.nextToken();y = (long)cin.nval;cin.nextToken();a = (long)cin.nval;long m = y / (x - 1);if(m == 0){//System.out.println("1");out.println("1");continue;}long g = gcd(a, m);long c = a / g;if(gcd(c, x) != 1){//System.out.println("Impossible!");out.println("Impossible!");continue;}long ans = Euler(c);int len = 0;for(int i = 1; i * i <= ans; i++){if(ans % i == 0){num[len++] = i;if(ans % i != i){num[len++] = ans / i;}}}Arrays.sort(num, 0, len);int i;for(i = 0; i < len; i++){if(erfen(x, num[i], c) == 1)break;}//System.out.println(num[i]);out.println(num[i]);}out.flush();}static long gcd(long a, long b){if(a < b){long temp = a;a = b;b = temp;}/*if(b == 0) return a;return gcd(b, a % b);*/while(b > 0){//a = b;long t = a % b;a = b;b = t;}return a;}static long Euler(long n){long sum = n;long temp = n;for(long i = 2; i * i <= n; i++){if(temp % i == 0){sum = sum / i * ( i - 1);while(temp % i == 0)temp /= i;}}if(temp > 1)sum = sum / temp * (temp - 1);return sum;}static long erfen(long a, long b, long mod){long temp = a % mod;long result = 1;while(b > 0){if(b % 2 == 1){result *= temp;result %= mod;b--;}else{temp *= temp;temp %= mod;b >>= 1;}}return result;}}