当返回值大于long时，如何从方法返回long？

Question

我想解决Topcoder中的一个问题，我需要在提供宽度和长度时找到矩形的数量（不包括正方形）。我的代码适用于所有测试用例，但对于最后一个测试用例，即592X964，我需要返回81508708664，这大于2^32-1，这大于long可容纳的值，但编译器要求必须返回值long。我已经添加了下面的问题陈述和我的代码，请好好理解一下，并让我知道是否有可能在java中使用long来返回上述值。

/* 
Problem Statement 
     
Given the width and height of a rectangular grid, return the total number of rectangles (NOT counting squares) that can be found on this grid. 
For example, width = 3, height = 3 (see diagram below): 
__ __ __ 
|__|__|__| 
|__|__|__| 
|__|__|__| 
In this grid, there are 4 2x3 rectangles, 6 1x3 rectangles and 12 1x2 rectangles. Thus there is a total of 4 + 6 + 12 = 22 rectangles. Note we don't count 1x1, 2x2 and 3x3 rectangles because they are squares. 
Definition 
     
Class: 
RectangularGrid 
Method: 
countRectangles 
Parameters: 
int, int 
Returns: 
long 
Method signature: 
long countRectangles(int width, int height) 
(be sure your method is public) 
Limits 
     
Time limit (s): 
2.000 
Memory limit (MB): 
64 
Notes 
- 
rectangles with equals sides (squares) should not be counted. 
Constraints 
- 
width and height will be between 1 and 1000 inclusive. 
Examples 
0) 

     
3 
3 
Returns: 22 
See above 
1) 

     
5 
2 
Returns: 31 
__ __ __ __ __ 
|__|__|__|__|__| 
|__|__|__|__|__| 
In this grid, there is one 2x5 rectangle, 2 2x4 rectangles, 2 1x5 rectangles, 3 2x3 rectangles, 4 1x4 rectangles, 6 1x3 rectangles and 13 1x2 rectangles. Thus there is a total of 1 + 2 + 2 + 3 + 4 + 6 + 13 = 31 rectangles. 
2) 

     
10 
10 
Returns: 2640 

3) 

     
1 
1 
Returns: 0 

4) 

     
592 
964 
Returns: 81508708664 

This problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited. (c)2003, TopCoder, Inc. All rights reserved. 
*/ 
public class RectangularGrid 
{ 
    public static void main(String[] args) 
    { 
     System.out.println(countRectangles(592,964)); 
    } 

    public static long countRectangles(int m, int n) 
    { 
     long tot_rect = (long)(((m*m)+m)*((n*n)+n))/4; 
     long tot_square = 0; 
     boolean status = true; 

     while(status) 
     { 
      if(m>0 && n>0) 
      { 
       tot_square+=(long)m*n; 
       --m; 
       --n; 
      } 
      else 
      { 
       status = false; 
      } 

     } 


     return tot_rect-tot_square; 
    } 
} 

Answer 1

运行此代码返回值的垃圾，我看不出为什么它是由长

如果该值可以容纳因为你正在做的int乘法：

public static long countRectangles(int m, int n) 
// Note ---------------------------^^^----^^^ 
{ 
    long tot_rect = (long)(((m*m)+m)*((n*n)+n))/4; 
    // Note -----------------^^^-------^^^ 

如果你想要那些使用long，你需要演员表：

public static long countRectangles(int m, int n) 
{ 
    long tot_rect = ((((long)m*m)+m)*(((long)n*n)+n))/4; 
    // Note -----------^^^^^^----------^^^^^^ 

施放最终结果对于中间结果中的损失没有任何作用。

Answer 2

长是8个字节，意思是它的范围是-2^63到2^63-1;所以它远远大于2^32-1。它是4字节的整型，范围是-2^31到2^31-1。所以，只要你在范围内，你就会很好。