By counting carefully it can be seen that a rectangular grid measuring 3 by 2 contains eighteen rectangles:
Although there exists no rectangular grid that contains exactly two million rectangles, find the area of the grid with the nearest solution.
如果数得足够仔细，能看出在一个3乘2的长方形网格中包含有18个不同大小的长方形，如下图所示：
尽管没有一个长方形网格中包含有恰好两百万个长方形，但有许多长方形网格中包含的长方形数目接近两百万，求其中最接近这一数目的长方形网格的面积
解题
有下面内容：
对于任意矩形M*N
其中1*1的矩阵有M*N个
1*2的矩阵有M*(N-1)个
2*1的矩阵有（M-1)*N个
实际上只要确定小矩阵左上角顶点在大矩形中的位置，这个矩阵的位置就唯一确定了
所有在任意矩形M*N中，矩阵i*j有（M-i+1)*(N-j+1)个
所以对于M*N的矩阵总的矩阵数量是：
int num = 0; for(int i =1;i<= m;i++){ for(int j =1;j<= n;j++){ num += (m-i + 1)*(n - j+1); } }
更让人想不到是是直接计算矩阵的数量：
num = （m+1）*m*(n+1)*n/4
Java
package Level3;import java.util.Random;public class PE085{ static void run() { int limit = 100; int close = Integer.MAX_VALUE; int area = 0; for(int m =1;m< limit ;m++){ for(int n = 1;n< limit ;n++){ int num = grid_num(m,n); if (num>2000000) break; if( Math.abs(num - 2000000 ) < Math.abs(close - 2000000)){ close = num; area = n*m; } } } System.out.println(area); } public static int grid_num2(int m , int n){ int num = 0; num = (m+1)*m*(n+1)*n/4; return num; }// 2772// running time=0s0ms public static int grid_num(int m , int n){ int num = 0; for(int i =1;i<= m;i++){ for(int j =1;j<= n;j++){ num += (m-i + 1)*(n - j+1); } } return num; }// 2772// running time=0s20ms public static void main(String[] args){ long t0 = System.currentTimeMillis(); run(); long t1 = System.currentTimeMillis(); long t = t1 - t0; System.out.println("running time="+t/1000+"s"+t%1000+"ms"); }}
你说是不是很流氓，这个规律，我怎么那么聪慧的会发现？
Python
# coding=gbkimport time as time t0 = time.time()def run(): limit = 100 close = 0 area = 0 for m in range(1,limit): for n in range(1,limit): num = grid_num(m,n) if num>2000000:break if abs(num - 2000000) < abs(close -2000000): close = num area = n*m print area def grid_num(m ,n): count = 0 for i in range(1,m+1): for j in range(1,n+1): count += (m-i+1)*(n-j+1) return count run()t1 = time.time()print "running time=",(t1-t0),"s"# 2772# running time= 1.19499993324 s