LA 4123 Glenbow Museum Castle Museum

Test instructions: For a polygon with an edge parallel to an axis, we can use a sequence of R and O to describe it: starting from a vertex in a counterclockwise order, hitting a 90° internal angle of R; touching a 270° internal angle. Such sequences are called angular sequences. Defining a star polygon is a point in a polygon where you can see each point on the polygon boundary. Now given a positive integer n, how many angular sequences of length l can correspond to at least one star polygon. where each edge of the polygon is arbitrarily long.

It seems very difficult to start, a little abstract. We can actually assume that the polygon length is 1 and then construct:

First, it's easy to see that the polygon vertex number is n, and the number of edges is n. And because the edges of the polygons are parallel to the axes, L is at least 4 and must be an even number. To make a polygon a star polygon, the polygon must have a core. In this problem. The inner angle of the polygon is only 90 ° and 270°, so it is clear that if there are two adjacent internal angles are 270°, then this polygon is not nuclear. That is, the angle sequence can not have two o adjacent, and the tail can not be at the same time O.

Then we determine the number of R and O. The internal angles of the N-edged shape are (N-2) *180, and R represents the inner angle of 90,o for 270, so it is easy to calculate the number of R and O, respectively (N + 4)/2 and (n-4)/2. The problem is then converted to how many methods of placing (n-4)/2 O and any two nonadjacent are placed in n positions.

This is a very simple combinatorial problem. The bundling method is a good idea for the non-adjacent problem. We bundle each o with an R and make R on the left of O. This is bundled into a whole, equivalent to placing in N-(n-4)/2 positions (n-4)/2 O, that is, placing 4 R. The answer is C ((n + 4)/2,4). Notice that a situation is missing, that is, the sequence begins with O. Because I just put R on the left of O. So we make the first one for O, so the last one must be r, the remaining position will still bind each O and R and let R on the left of O. The answer is C ((n + 2)/2,4). So the final answer is C ((n + 4)/2,4) + C ((n + 2)/2,4). Be careful not to forget that L is an odd or less than 4 case with a special output of 0.

#include <iostream> #include <algorithm> #include <cstring> #include <cmath> #include < cstdio> #include <iomanip> #include <string>using namespace Std;long long L;int main () {    int case = 0;    while (scanf ("%lld", &l) && L)    {        printf ("Case%d:", ++case);        if (l&1 | | L < 4)        {            printf ("0\n");            Continue;        }        Long Long ans1 = (l + 4) >> 1, Ans2 = (l + 2) >> 1;        ans1 = ans1* (ans1-1) * (ans1-2) * (ans1-3)/24;        Ans2 = ans2* (ans2-1) * (ans2-2) * (ans2-3)/24;        printf ("%lld\n", ans1 + ans2);    }    return 0;}



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LA 4123 Glenbow Museum Castle Museum