10个8g的二进制文件，其中每8个字节为一个64bits整数，对这10个文件进行快速排序处理，找出10个文件中重复的数据，使用java语言实现上述操作，要求输出重复数据的整数值，该值所在的文件，以及所在文件的位置，并将输出结果存储到Task3.out文件中

好的，以下是使用Java语言实现上述操作的一种可能的解决方案：

import java.io.*;
import java.nio.ByteBuffer;
import java.nio.LongBuffer;
import java.nio.channels.FileChannel;
import java.util.Arrays;

public class SortAndFindDuplicates {

    private static final int LONG_SIZE = 8; // 8 bytes per long
    private static final int BUFFER_SIZE = 1024 * 1024; // 1MB buffer

    private static void sortFile(String filename) throws IOException {
        RandomAccessFile file = new RandomAccessFile(filename, "rw");
        FileChannel channel = file.getChannel();
        ByteBuffer buffer = ByteBuffer.allocate(BUFFER_SIZE);
        LongBuffer longBuffer = buffer.asLongBuffer();
        long[] array = new long[BUFFER_SIZE / LONG_SIZE];
        int bytesRead;
        do {
            buffer.clear();
            bytesRead = channel.read(buffer);
            if (bytesRead > 0) {
                longBuffer.flip();
                longBuffer.get(array, 0, bytesRead / LONG_SIZE);
                Arrays.sort(array, 0, bytesRead / LONG_SIZE);
                longBuffer.clear();
                longBuffer.put(array, 0, bytesRead / LONG_SIZE);
                channel.position(channel.position() - bytesRead);
                buffer.limit(bytesRead);
                channel.write(buffer);
            }
        } while (bytesRead > 0);
        channel.close();
        file.close();
    }

    private static void findDuplicates(String[] filenames, String outputFilename) throws IOException {
        int numFiles = filenames.length;
        RandomAccessFile[] files = new RandomAccessFile[numFiles];
        FileChannel[] channels = new FileChannel[numFiles];
        ByteBuffer[] buffers = new ByteBuffer[numFiles];
        LongBuffer[] longBuffers = new LongBuffer[numFiles];
        long[][] arrays = new long[numFiles][BUFFER_SIZE / LONG_SIZE];
        int[] positions = new int[numFiles];
        for (int i = 0; i < numFiles; i++) {
            files[i] = new RandomAccessFile(filenames[i], "r");
            channels[i] = files[i].getChannel();
            buffers[i] = ByteBuffer.allocate(BUFFER_SIZE);
            longBuffers[i] = buffers[i].asLongBuffer();
            positions[i] = -1;
        }
        int bytesRead;
        boolean finished = false;
        while (!finished) {
            for (int i = 0; i < numFiles; i++) {
                if (positions[i] == -1) {
                    bytesRead = channels[i].read(buffers[i]);
                    if (bytesRead == -1) {
                        finished = true;
                        break;
                    }
                    buffers[i].flip();
                    longBuffers[i].limit(bytesRead / LONG_SIZE);
                    longBuffers[i].get(arrays[i], 0, bytesRead / LONG_SIZE);
                    buffers[i].clear();
                    positions[i] = 0;
                }
            }
            if (!finished) {
                long minValue = Long.MAX_VALUE;
                int minIndex = -1;
                for (int i = 0; i < numFiles; i++) {
                    if (positions[i] != -1 && arrays[i][positions[i]] < minValue) {
                        minValue = arrays[i][positions[i]];
                        minIndex = i;
                    }
                }
                int count = 0;
                for (int i = 0; i < numFiles; i++) {
                    if (positions[i] != -1 && arrays[i][positions[i]] == minValue) {
                        count++;
                    }
                }
                if (count > 1) {
                    try (BufferedWriter writer = new BufferedWriter(new FileWriter(outputFilename, true))) {
                        writer.write(String.format("%d found in %s, positions: ", minValue, filenames[minIndex]));
                        for (int i = 0; i < numFiles; i++) {
                            if (positions[i] != -1 && arrays[i][positions[i]] == minValue) {
                                writer.write(String.format("%d ", positions[i]));
                            }
                        }
                        writer.newLine();
                    }
                }
                for (int i = 0; i < numFiles; i++) {
                    if (positions[i] != -1 && arrays[i][positions[i]] == minValue) {
                        positions[i]++;
                        if (positions[i] == BUFFER_SIZE / LONG_SIZE || positions[i] == longBuffers[i].limit()) {
                            positions[i] = -1;
                        }
                    }
                }
            }
        }
        for (int i = 0; i < numFiles; i++) {
            channels[i].close();
            files[i].close();
        }
    }

    public static void main(String[] args) throws IOException {
        String[] filenames = {"file1.bin", "file2.bin", "file3.bin", "file4.bin", "file5.bin",
                "file6.bin", "file7.bin", "file8.bin", "file9.bin", "file10.bin"};
        for (String filename : filenames) {
            sortFile(filename);
        }
        findDuplicates(filenames, "Task3.out");
    }

}

该程序将首先对每个文件进行快速排序，然后使用归并排序的思想在所有文件中查找重复项。对于每个重复项，程序将其整数值、文件名和位置写入Task3.out文件中。

请注意，该程序假定每个文件的大小不超过2GB，因为Java中的数组大小限制为Integer.MAX_VALUE。如果您的文件更大，请考虑使用外部排序或其他技术来处理文件。