octagon geometry

[MAXM]; to DoubleDX[MAXN]; + DoubleDY[MAXN]; - DoubleDIS[MAXN]; the intMain () * { $ intCnt=0;Panax Notoginseng while(~SCANF ("%d%d",n,m)) - { the Doublesum=0; +node[0].x=0; Anode[0].y=0; thea[0]=0; + for(intI=1; i) - { $scanf"%LF%LF",node[i].x,node[i].y); $dx[i]=node[i].x-node[i-1].x; -dy[i]=node[i].y-node[i-1].y; -Dis[i]=sqrt ((node[i].x-node[i-1].x) * (node[i].x-node[i-1].x) + (node[i].y-node[i-1].Y) * (node[i].y-node[i-1].y)); thesum+=Dis[i]; -Dx[i]/

Title Address: POJ 1265Test instructions: Given a lattice polygon, find the internal number in, the number of points on the edge, and the area S.Ideas: The use of a lot of theorems.1. Pique theorem: s=in+on/2-1, i.e. in= (2*s+2-on)/2.2. Polygon Area formula: The sum of the cross product of a vector consisting of two adjacent points and the origin is calculated sequentially.3. Find the number of lattice points on the edge: a segment that is vertex-based, with points that are covered by GCD (Dx,dy

of it center and radius, expressed as real numbers in decimal nota tion, with up to digits after the decimal point. The imprecision margin is 5 x 10-13 . That's, it is guaranteed this variations of less than 5 x 10-13 on input values does not change which dis CS is visible. Coordinates of all points contained in discs is between-10 and 10.Confetti is listed in their stacking order, x1y1r1 being the Bottom one and xnynrN the top one. You is observing from the top.The end of the input is marked

Practice your code skills, complete the projecteuler topic, and make an archival record.Question 1th:If we list all the natural numbers below is multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.Find the sum of all the multiples of 3 or 5 below 1000.Find below 1000 and divide the numbers by 3 or 5 and calculate their and.def multiples_of_3_and_5 (num): sum = 0 for n in xrange (num): If isn't n%3 or not n%5:sum+=n Retu RN sumif __name__ = = "__main__": Print multiples_of

Practice your code skills, complete the projecteuler topic, and make an archival record.Question 1th:If we list all the natural numbers below is multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.Find the sum of all the multiples of 3 or 5 below 1000.Find below 1000 and divide the numbers by 3 or 5 and calculate their and.def multiples_of_3_and_5 (num): sum = 0 for n in xrange (num): If isn't n%3 or not n%5:sum+=n Retu RN sumif __name__ = = "__main__": Print multiples_of

format "s = L", where S is the name of the unknown side (a, B or c), and L are its length. L must is printed exact to three digits to the "right" of the decimal point.Print a blank line after each test case.Sample Input3 4-1-1 2 75-1 30 0 0Sample OutputTriangle #1c = 5.000Triangle #2A = 6.708Triangle #3Impossible.At the beginning of the operation due to the lack of consideration to consider some of the circumstances of the two caused errors, such as one of the a,c,b in the case of 0, as well as

POJ 1696Judging the cross product, in judging the angle.Like a convex hull, but without a convex hull version.So I thought of a way.POJ 2074Handle each obstacle in the middle of the projection on the road, and then scan.Pay attention to both ends of the situation!Special thanks for the data in discussPOJ 1654Polygon AreaPOJ 1410My practice is to force judgment, line intersection + Special case (segment in the rectangle inside)Thanks again for the data in discussAlso, must learn English, see the

The problem is asking you to ask for a line that can pass through the most fruit (a point is counted).It can be proved that it is possible to enumerate the lines consisting of two points.Because if there is a line that crosses n fruits, it is necessary to translate some of it so that it passes through n but it can no longer be translated, so the line must be at some end of a certain fruit.Again with this end, rotate this line, or through N, until it cannot be rotated (then the rotation may not p

single line of output should appear. This line should take the form of:"N:this property would begin eroding in year Z."Where N is the data set (counting from 1), and Z are the first year (start from 1) This property would be within the SEMICIR CLE at the END of year Z. Z must is an integer.After the last data set, this should print out "END of OUTPUT."Notes:1. No property would appear exactly on the semicircle boundary:it would either be inside or outside.2. This problem'll be judged automatica

Turn the CornerTime limit:3000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 2229 Accepted Submission (s): 856Problem descriptionmr. West bought a new car! So he's travelling around the city.One day he comes to a vertical corner. The street he is currently in have a width x, the street he wants to turn to have a width y. The car has a length l and a width d.Can Mr West go across the corner?Inputevery line have four real numbers, X, Y, L and W.Proceed to th

Chapter Three, the local theory of the surface1. The concept of surfaces1.1. The concept of surfaces1.2. Tangent plane and FA2. The first basic form of the surface3. Second basic form of the surface4. Normal curvature and Weingarten transformation5. Main curvature and Gauss curvature6. Some examples of surfaces6.1. Rotating surfaces6.1.1. Gauss curvature Rotational surface6.1.2. Constant average curvature rotational surface6.2. Ruled surface and developable surfaces6.3. Full umbilical point surf

= sqrt ((a.x-b.x) * (a.x-b.x) + (A.Y-B.Y) * (A.Y-B.Y) + (a.z-b.z) * (a.z-b.z)); returnans;}BOOLJudge (Point A, point B, point C, point D)///determine whether the four points are coplanar, and the Coplanar return is true;{point S1, S2, S3; s1.x= b.x-a.x; S1.y = B.Y-A.Y; S1.z = b.z-a.z; s2.x= c.x-a.x; S2.y = C.Y-A.Y; S2.z = c.z-a.z; s3.x= d.x-a.x; S3.y = D.Y-A.Y; S3.z = D.z-a.z; intAns = s1.x*s2.y*s3.z + s1.y*s2.z*s3.x + s1.z*s2.x*s3.y-s1.z*s2.y*s3.x-s1.x*s2.z*s3.y-s1.y*s2.x*s3.z; returnAns = =0;

Click Open Link Miaomiao's geometry Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/65536 K (Java/Others)Total submission (s): 438 accepted submission (s): 107 Problem descriptionthere are n point on X-axis. miaomiao wowould like to cover them all by using segments with same length. There are 2 limits: 1. A point is convered if there is a segments t, the point is the left end or the right end of T. 2. The length of the intersection of an

HDU 4932 Miaomiao #39; s Geometry Brute ForceClick Open LinkMiaomiao's GeometryTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission (s): 438 Accepted Submission (s): 107Problem DescriptionThere are N point on X-axis. Miaomiao wowould like to cover them ALL by using segments with same length.There are 2 limits:1. A point is convered if there is a segments T, the point is the left end or the right end of T.2.

HDU 1086 You can Solve a Geometry Problem too (determine the intersection of line segments) Address: HDU 1086 In this case, I wrote more than 2 k B code to determine the intersection of line segments .. Is it a waste... But I don't think I can optimize it any more .... The cross product is used to determine the intersection of line segments. If the two line segments are L1 and L2, calculate the cross product of the vectors of one of L1 and L2 endpoint

http://www.lydsy.com/JudgeOnline/problem.php?id=1043The only thing that I don't want is how to find the circumference of the circle and Qaaq ...and find the Good God! We can turn the arc into $[0, 2 \pi]$ line!Then be sure to pay attention! Starting point is $ (1, 0) $ (unit circle)First I learned the cosine theorem ...In the triangle ABC$ $cos a=\frac{| ab|^2+| ac|^2-| bc|^2}{2| ab| | ac|} $$Proving very simple ...$$\begin{align}| {Bc}|^2 = \VEC{BC} \cdot \VEC{BC} \ \ = (\vec{ac}-\vec{ab}) \cd

http://poj.org/problem?id=1556First, each line of the path must be a connection between the endpoints. Prove? It's a pit. Anyway, I did a random painting and then I wrote it.And what is the rhythm of re? I remember I had enough to drive ... And then open the big point just a ... It's so embarrassing.#include 　　 DescriptionYou is to find the length of the shortest path through a chamber containing obstructing walls. The chamber always has sides at x = 0, x = ten, y = 0, and y = 10. The init

(SCANF ("%D%LF", n,r)!=eof n+R) { for(i=1; i) {scanf ("%LF%LF", c[i].c.x,c[i].c.y), C[I].R =2.0*R; Sc[i]= C[i], SC[I].R =R; } Vectorsec; Sec.clear (); for(i=1; i) { for(j=i+1; j) getcirclecircleintersection (C[I],C[J],SEC); } DoubleMini =Mod; if(Check (Point (0,0)) {printf ("%.6f\n",0.0);Continue; } for(i=1; i) { if(DCMP (DISP (c[i].c)) = =0) { if(Check (Point (2*r,0)) Mini = min (Mini,2*R); Continue; } sec.push_back (Point (c[i].c+c[i].c* (

,1],a[i,2]); -Sort1, n); j:=1; Wu fori:=2 toN Do//Go heavy - begin About if(A[i,1]1])or(A[i,2]2]) Then $ begin - Inc (J); -A[j,1]:=a[i,1];a[j,2]:=a[i,2]; - End; A End; +n:=J; the//Convex bag - fori:=1 toN Dod[i]:=I;doit (B,M1); $ fori:=1 toN Dod[i]:=n+1-I;doit (c,m2); the//two halves of integration the fori:=1 toM1 Dod[i]:=B[i]; the fori:=2 toM2 DoD[i+m1-1]:=C[i]; theStart calculating perimeter +

Step one to draw the point. Select the Point tool on the left toolbox, draw two points A and B in the vertical direction, and then draw two points C and D in the same horizontal direction as the point A, as shown in the following figure; Draw a four-point example with a point tool in a geometry artboard Step two to paste the picture. Select B, c two points, the mouse click on the top of the "Edit" menu, in its drop-down op