geometry for idiots

The main idea: N points on a given plane, to find the minimum of one point in the N point to the Chebyshev distance of the n pointsChebyshev, that is, the maximum of the absolute value of each coordinate differenceFirst of all, if we want to convert the Manhattan distance to Chebyshev distance then we have to turn the point (x, y) into a new point (x+y,x-y) where the Chebyshev distance between the points is the Manhattan distance between the origin.Similarly, we can convert the Chebyshev distanc

[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