挑战程序竞赛系列（15）：2.6快速幂运算

详细代码可以fork下Github上leetcode项目，不定期更新。

练习题如下：

• POJ 3641: Pseudoprime numbers
• POJ 1995: Raising Modulo Numbers

POJ 3641: Pseudoprime numbers

判断在当前基数a时，满足费马小定理的伪素数。

代码如下：

public static void main(String[] args) throws IOException {        Scanner in = new Scanner(System.in);        while (true) {            int p = in.nextInt();            int a = in.nextInt();            if (p == 0 && a == 0)                break;            if (a != quickPow(a, p, p)){                System.out.println("no");                continue;            }            if (isPrime(p)) System.out.println("no");            else System.out.println("yes");        }    }    public static boolean isPrime(int num){        for (int i = 2; i * i <= num; ++i){            if (num % i == 0) return false;        }        return true;    }    public static long quickMul(long a, long b, long mod) {        long ans = 0;        while (b != 0) {            if ((b & 1) != 0) {                ans = (ans + a) % mod;            }            b >>= 1;            a = (a + a) % mod;        }        return ans;    }    public static long quickPow(long a, long b, long mod) {        long ans = 1;        while (b != 0) {            if ((b & 1) != 0) {                ans = quickMul(ans, a, mod);            }            b >>= 1;            a = quickMul(a, a, mod);        }        return ans;    }

POJ 1995: Raising Modulo Numbers

取模运算，水题。

代码如下：

public static void main(String[] args) throws IOException {        Scanner in = new Scanner(System.in);        int Z = in.nextInt();        while ((Z--) != 0){            int M = in.nextInt();            int N = in.nextInt();            long ans = 0;            for (int i = 0; i < N; ++i){                long a = in.nextLong();                long b = in.nextLong();                ans = (ans + quickPow(a, b, M)) % M;            }            System.out.println(ans);        }    }    public static long quickMul(long a, long b, long mod){        long res = 0;        while (b != 0){            if ((b & 1) != 0){                res = (res + a) % mod;            }            b >>= 1;            a = (a + a) % mod;        }        return res;    }    public static long quickPow(long a, long b, long mod){        long res = 1;        while (b != 0){            if ((b & 1) != 0){                res = quickMul(res, a, mod);            }            b >>= 1;            a = quickMul(a, a, mod);        }        return res;    }