Idea 1: treat all squares as two types and divide them by height. One is called Class A, and the height is the unit length. The other is called Class B, and its height is half the diagonal length of a unit square. Look at Class A first, in which the height can only be m kinds of heights from 1 to m. When the height is n, I set k=m-n+ 1. Then, in the rectangle of n*m, there are k*(2k- 1) squares, and in the whole graph of m * m, there are 2k n *. In addition, the small square of n*n is counted again (because there are four squares in the 2k rectangle of n*m * * * k, so the number of squares with high n is k * (2k. Therefore, the number of class A squares is the sum of k from 1 to m&((4k-3) * k 2); In this case, it is from 1 to 7. In fact, if you are familiar with the cube formula, it is easy to calculate the sum of 4 * k 3-3 * k 2. When m=7,&; A=27 16
Looking at Class B again, through the analysis of Class A, we can see that all figures such as trapezoid, parallelogram, square and rectangle appear in Class A, and Class B only contains one figure, namely isosceles trapezoid, with the diagonal of the original figure as two bases, and this trapezoid must take the positive diagonal of a small square as the long base. We know that there are n*n units of k 2 squares in the square of m*m, 2*(n- 1) isosceles trapeziums with positive diagonal as long base in the square of n * m, and the isosceles trapeziums contained in the square of m * m are the sum of n from 1 to m&; 2 * (n-1) * k 2, when m=7, & B=392, so
& ampa+& amp； B=27 16+392=3 108。
It sounds complicated, but it still works. Clear your mind.
Idea 2: I just think of all the squares as triangular saws, without going deep.
I don't know if I made it clear. Welcome to discuss.
Answer supplement: Square is regarded as two types in the train of thought 1, which is convenient for imagination and calculation and is not an accurate description. In the horizontal direction considered by Class A, Class B is 45 degrees. That rhombic parallelogram is placed in class A when calculating, so there is only isosceles trapezoid in class B.
Besides, I set K just for writing.
As for the n*m rectangle in the m*m square, if there are n squares in the Y-axis direction, every time a grid moves up, it is a new n*m rectangle, so there are m-n+ 1 long-bottom rectangles, and m-n+ 1 short-bottom rectangles in the vertical direction are similar to the horizontal direction, so there are 2*k such rectangles.
Another problem is the k*(2k- 1) quadrilateral in this kind of rectangle. Because quadrangles are divided by height, only n quadrangles with unit height are considered in an n*m rectangle. For example, in an n*m rectangle at the lowest level, there are 2k- 1 quadrangles with upper, lower and rightmost sides respectively (this is because the straight line that determines this height is only 2(n-m)=2(k+ 1) (there are n-m related points, and each point is connected by two lines, so it is 2 * (. It means that (from the rectangle of 1*m, gradually increase to n*m, and then increase to m*m) the height in front has been calculated, so there is 2k- 1). Just change the line on the left to the right, and the upper and lower sides will be fixed. The quadrilateral found is 2k-2, and so on, * * is 2k *. Descriptive words are too much trouble. Just give an example or count. Ha ha.
There is no reference list, and there is no need for principles or theorems, websites, etc.
Specific analysis: 7*7 square.
The rectangles are 1*7, 2*7, 3*7, 4*7, 5*7, 6*7 and 7*7. The 7*7 model has 14 different 1*7 models, 12 2*7 models and so on. It should be no problem.
Analyzing a 2*7 type, these squares are similar in some ways. So analyze one casually, just look at the bottom 7*2, and don't cross out the others. Consider a square with a height of 2. The horizontal line in the middle and the two small diagonals on both sides are useless. Because these two paragraphs are squares with a height of 1, they are not considered. Therefore, all useful lines have 8 lines and 6 diagonals perpendicular to the bottom and upper and lower boundaries. Let's number the points, point a 1 a2...a8, and point b 1 b2 ... b8.
Fix a 1b 1 and find a quadrilateral (like a 1b 1b2a2 is a square with a height of 1 and should not be counted in the rectangle of 1*7), so there is a1. A1b1b4a ... a1b1b8a8, * * 1 1, and then fix a 1b2 to find quadrilaterals, including a/kloc-0. This adds up to the so-called k (2k-1) = 6 *11= 66.
Is that clear? Hehe.