java多线程协作 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。并输出结果

好的，这个问题可以回答。以下是一个可能的解决方案：

public class MultiThreadSum {
    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 InterruptedException {
        long[] results = new long[THREAD_COUNT];
        Thread[] threads = new Thread[THREAD_COUNT];

        for (int i = 0; i < THREAD_COUNT; i++) {
            final int index = i;
            threads[i] = new Thread(() -> {
                long start = START + (END - START) / THREAD_COUNT * index;
                long end = START + (END - START) / THREAD_COUNT * (index + 1);
                if (index == THREAD_COUNT - 1) {
                    end = END + 1;
                }
                long sum = 0;
                for (long i1 = start; i1 < end; i1++) {
                    sum += i1;
                }
                results[index] = sum;
            });
            threads[i].start();
        }

        for (Thread thread : threads) {
            thread.join();
        }

        long total = 0;
        for (long result : results) {
            total += result;
        }
        System.out.println("Total sum: " + total);
    }
}

这个程序创建了6个线程，每个线程计算一部分数字的总和，最后把结果加起来得到总和。每个线程计算的数字量之差不超过1，确保了计算的正确性。