Question

我发现了类似的问题但不是我想要的问题。

我需要将矩阵1024x1024分成相等的4个部分，但我必须花费每个部分的开头和结尾来处理另一个带线程的类。我想过使用两个循环，但我不知道如何正确更改变量的值，因为它超过了最大值。

代码：

private static final int MAX_MATRIX = 1024;
private static final int NUM_THREADS = 4;
int start_row = 0, end_row = 0, start_column = 0, end_column = 0;
for (int i = 0; i < NUM_THREADS/2; i++){
 end_row = start_row + (MAX_MATRIX/2);      
 for (int j = 0; j < (NUM_THREADS/2); j++){
    end_column = start_column + (MAX_MATRIX/2);
    myTask[i] = new TaskAnalyze(start_row, end_row, start_column, end_column, int_radom);
    myThreads[i] = new Thread(myTask[i],"TASK " + i);
 }      
 start_column += MAX_MATRIX/2;      
 start_row += MAX_MATRIX/2; 
}

int_random是矩阵。

应该是：

myTask [0] = new TaskAnalyze（0,512,0,512，int_radom）;

myTask [1] = new TaskAnalyze（0,512,512,1024，int_radom）;

myTask [2] = new TaskAnalyze（512,1024,0,512，int_radom）;

myTask [3] = new TaskAnalyze（512,1024,512,1024，int_radom）;

有什么建议吗？

Answer 1

试试这个：

int MAX_MATRIX = 1024;
int NUM_THREADS = 4;
int start_row = 0, end_row = 0, start_column = 0, end_column = 0;
for (int i = 0; i < NUM_THREADS/2; i++){
    start_row = i*MAX_MATRIX/2;
    end_row = start_row + MAX_MATRIX/2;
    for (int j = 0; j < NUM_THREADS/2; j++){
        start_column = j*MAX_MATRIX/2;
        end_column = start_column + MAX_MATRIX/2;
        System.out.println(start_row + "," + end_row + "," + start_column + "," + end_column);
    }
}

应该给你以下输出：

0,512,0,512
0,512,512,1024
512,1024,0,512
512,1024,512,1024

然后你可以用你的行替换System.out.println行：

myTask[i] = new TaskAnalyze(start_row, end_row, start_column, end_column, int_radom);
myThreads[i] = new Thread(myTask[i],"TASK " + i);

在java中拆分矩阵

1 个答案: