SRM146 DIV1 300

数学题，分析如下：

宽度w、高度h的矩形（包括正方形）在宽度width、高度height的矩形中所含个数是：$(width-w+1)(height-h+1)$

因此所求不等边矩形个数：

$\begin{equation} \begin{split}  sum &= \sum_{w=1}^{width} \sum_{h=1}^{height} (width-w+1)(height-h+1) - \sum_{w=1}^{min(width,height)}(width-w+1)(height-w+1) \\ &= \sum_{w=1}^{width} \sum_{h=1}^{height} width*height - \sum_{w=1}^{min(width,height)}(width-w+1)(height-w+1) \\ &= (1+width)width/2*(1+height)height/2 - \underline{\sum_{w=1}^{min(width,height)}(width-w+1)(height-w+1))} \end{split} \end{equation}$

注意下划线所指式子要用程序循环累加来计算

class RectangularGrid:
    def countRectangles(self, width, height):
        mn = min(width, height)
        return int((1+width)*width/2) * int((1+height)*height/2) - countSquares(width,height)


def countSquares(width, height):
    sum = 0
    k = min(width, height)
    for x in range(1, k+1):
        sum += (width-x+1) * (height-x+1)
    return sum


# test
o = RectangularGrid()

# test case
assert(o.countRectangles(3,3) == 22)
assert(o.countRectangles(5,2) == 31)
assert(o.countRectangles(10,10) == 2640)
assert(o.countRectangles(1,1) == 0)
assert(o.countRectangles(592,964) == 81508708664)

PS：好多人直接用O(width*height)的算法累加来计算的，比赛中为了编码速度，应该是可以接受的吧