**(1) A maximum of n straight lines can be split into planes.**

** The question is roughly as follows: n straight lines. A maximum of n areas can be divided into a plane.**

** Analysis: You may have seen this question before. This is at best a junior high school question. However, a type of question can be easily discovered only by starting with a simple one. When there are n-1 lines, the plane can be divided into a maximum of F (n-1) areas. If the n-th line has the largest number of cut-in areas, it must intersection each line and cannot have the same intersection. In this way, n-1 intersections are obtained. These links divide the nth line into two rays and The N-2 line is broken. Each Ray and line break will be split into two parts in some areas. So there are 2 + (n-2) more areas.**

** ** **So: F (n) = f (n-1) + n**

** = F (n-2) + (n-1) + n**

** ......**

** = F (1) + 1 + 2 + ...... + N**

** = N (n + 1)/2 + 1**

** (2) line split plane (hdu2050)**

** According to the line split plane, the intersection determines the number of rays and line segments, and then determines the number of new regions. When n-1 lines are broken, the number of areas is F (n-1 ). To maximize the number of added areas, the line segments on both sides of the line must be at the same time as the n-1 line edges, that is, 2 * (n-1) line segments. The number of new lines is 4 * (n-1) and the number of rays is 2. Note that only one area can be added to the adjacent two line segments.**

** **

** Therefore, F (n) = f (n-1) + 4 (n-1) + 2-1**

** = F (n-1) + 4 (n-1) + 1**

** = F (n-2) + 4 (n-2) + 4 (n-1) + 2**

** ......**

** = F (1) + 4 + 4*2 + ...... + 4 (n-1) + (n-1) **

** = 2n ^ 2-N + 1**

** (3) Plane Separation of Closed Curves**

** For example, if n closed curves are arranged on a plane, and any two closed curves exactly intersect at two points, and any three closed curves do not intersect at the same point, ask the number of areas in which these closed curves divide the plane.**

** Analysis: When n-1 circles, the number of regions is F (n-1 ). then, the nth circle must be the first n-1 circle. Then, the nth circle is divided into 2 (n-1) segments, and 2 (n-1) areas are added.**

** **

** So: F (n) = f (n-1) + 2 (n-1)**

** = F (1) + 2 + 4 + ...... + 2 (n-1)**

** = N ^ 2-N + 2**

** (4) plane split space problem (hdu1290)**

** From the two-dimensional segmentation problem, we can see that the plane segmentation is related to the intersection of lines, that is, the intersection determines the number of rays and line segments, thus determining the number of new regions. In 3D, is it related to the plane line? When there are n-1 planes, the number of split spaces is F (n-1 ). To have the maximum number of spaces, the n-th plane must intersection the first n-1 plane and cannot have a common intersection. That is, up to n-1**

Line Break. The N-1 line splits the n-th plane into a maximum of g (n-1) areas. (G (n) is the number of straight line split planes in (1) This plane splits the original space into two, then a maximum of N-1-1 space is added.

** **

** Therefore, F = f (n-1) + g (n-1) PS: G (n) = n (n + 1)/2 + 1**

** = F (n-2) + g (n-2) + g (n-1)**

** ......**

** = F (1) + G (1) + G (2) + ...... + G (n-1)**

** = 2 + (1*2 + 2*3 + 3*4 + ...... + (N-1) N)/2 + (n-1)**

** = (1 + 2 ^ 2 + 3 ^ 2 + 4 ^ 2 + ...... + N ^ 2-1-2-3 -...... -N)/2 + n + 1**

** = (N ^ 3 + 5N)/6 + 1**