用java分别实现以下两个程序(1) 写两个线程，一个线程打印1——52，另一个线程打印A——Z，打印顺序为：12A34B...5152Z (2) 编程实现三线程程序，其中一个线程计算1-1000的累加和，另一个线程求100！，在主线程输出1-1000的累加和与100！。

(1)

public class PrintThread implements Runnable {

    private int count = 1;
    private char ch = 'A';

    @Override
    public void run() {
        synchronized (this) {
            while (count <= 52 && ch <= 'Z') {
                if (count % 2 == 1) {
                    System.out.print(count++);
                    System.out.print(count++);
                } else {
                    System.out.print(ch++);
                }
            }
        }
    }
}

public class Main {
    public static void main(String[] args) {
        PrintThread printThread = new PrintThread();
        new Thread(printThread).start();
    }
}

(2)

public class SumThread implements Runnable {

    private int sum = 0;

    @Override
    public void run() {
        for (int i = 1; i <= 1000; i++) {
            sum += i;
        }
    }

    public int getSum() {
        return sum;
    }
}

public class FactorialThread implements Runnable {

    private BigInteger factorial = BigInteger.ONE;

    @Override
    public void run() {
        for (int i = 2; i <= 100; i++) {
            factorial = factorial.multiply(BigInteger.valueOf(i));
        }
    }

    public BigInteger getFactorial() {
        return factorial;
    }
}

public class Main {
    public static void main(String[] args) {
        SumThread sumThread = new SumThread();
        FactorialThread factorialThread = new FactorialThread();
        Thread thread1 = new Thread(sumThread);
        Thread thread2 = new Thread(factorialThread);
        thread1.start();
        thread2.start();
        try {
            thread1.join();
            thread2.join();
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
        System.out.println("1-1000的累加和为：" + sumThread.getSum());
        System.out.println("100!的值为：" + factorialThread.getFactorial());
    }
}