Java实现分成6个线程，计算m到n的值(以1到100000000为例)的总和。要求每个线程计算的数字量之差不超过1。

好的，这是一个多线程编程问题。可以使用Java的Executor框架来实现。以下是一个可能的解决方案：

import java.util.concurrent.*;

public class SumCalculator {
    private static final int THREAD_COUNT = 6;
    private static final long START = 1;
    private static final long END = 100000000;

    public static void main(String[] args) throws Exception {
        ExecutorService executor = Executors.newFixedThreadPool(THREAD_COUNT);
        long step = (END - START + 1) / THREAD_COUNT;
        long remainder = (END - START + 1) % THREAD_COUNT;
        long[] starts = new long[THREAD_COUNT];
        long[] ends = new long[THREAD_COUNT];
        for (int i = 0; i < THREAD_COUNT; i++) {
            starts[i] = START + i * step;
            ends[i] = starts[i] + step - 1;
            if (i < remainder) {
                ends[i]++;
            }
            executor.submit(new SumTask(starts[i], ends[i]));
        }
        executor.shutdown();
        executor.awaitTermination(1, TimeUnit.HOURS);
        System.out.println("Total sum: " + SumTask.totalSum);
    }

    private static class SumTask implements Runnable {
        private final long start;
        private final long end;
        private static long totalSum = 0;

        public SumTask(long start, long end) {
            this.start = start;
            this.end = end;
        }

        @Override
        public void run() {
            long sum = 0;
            for (long i = start; i <= end; i++) {
                sum += i;
            }
            synchronized (SumTask.class) {
                totalSum += sum;
            }
        }
    }
}

这个程序使用了一个Executor框架来创建6个线程，每个线程计算一部分数字的总和。为了确保每个线程计算的数字量之差不超过1，我们首先计算出每个线程应该计算的数字范围，然后将余数分配给前几个线程。每个线程计算完自己的数字范围后，将结果加入一个共享的总和变量中。最后，我们等待所有线程完成，输出总和。