topcoder arena

); } }returnAns }};1000 points:Dp[i][sta][k] Represents the enumeration to the I card, /************************************************************************* > File name:1000.cpp > Author: ALex > Mail: [email protected] > Created time:2015 May 08 Friday 21:25 26 seconds *********************************** *************************************/#include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespac

http://community.topcoder.com/stat?c=problem_statementpm=13625rd=16278First of all, if you remember the question of the minimum distance in Manhattan, you will think of a one-dimensional case where the point appears to be the smallest in the median. It is also shown here that four points, which appear at the median position, are the smallest.The way to do this is to test all the things that minimize the absolute value of an idea.My code is a little ugly, but it's over:#include 　　[

].length (); j1++) { for(intI2 = I1; I2 for(intJ2 = J1; J2 0].length (); j2++) {inta1 = A[i2][j2];if(I1 >0) A1-= a[i1-1][J2];if(J1 >0) A1-= a[i2][j1-1];if(I1 >0 J1 >0) A1 + = a[i1-1][j1-1];if(A1 = = k) {intT1, t2, anst =0; T1 = i1, t2 = (int) b.size ()-I2-1; Anst + =2* min (t1, T2) + max (t1, T2); T1 = J1, t2 = (int) b[0].length ()-J2-1; Anst + =2* min (t1, T2) + max (t1, T2); ans = min (ans, anst); } } } } }if(ans = =1e

{ Public:stringShortestcommon (stringS1,stringS2) {intT1 =0, t2 =0; for(inti =0; I if(S1[i] = =' * ') T1 = i; for(inti =0; I if(S2[i] = =' * ') t2 = i;if(T1 > T2) swap (S1, S2);stringANS1 ="", ANS3 =""; for(inti =0; I if(S1[i] = =' * ') ans1 = S1.substr (0, i); for(inti =1; I if(s2[i-1] ==' * ') Ans3 = S2.substr (i, s2.length ()-i); while(s1[0] !=' * ' s2[0] !=' * ') {if(s1[0]! = s2[0])return "Impossible"; S1.erase (0,1), S2.erase (0,1); } while(S1.back ()! =' * ' s2.back ()! =' * ')

Topcoder SRM 643 DIV1 250ProblemGive an integer n, and then give a VectorLimitsTime Limit (MS): 2000Memory Limit (MB): 256N: [2,10^18]Solutionn is continuously removed from the number of V (N/=v[i];) and a new n is obtained, which is recorded as N1. The N1 is then decomposed factorization. In [1,10^6] scanning, will N1 decomposition, to obtain a new N1, recorded as N2. If the N2 is not 1, it can be proved that N2 must be a prime number, add ans. The A

Test instructions: Give two large integers to determine which one is larger. Large integers are given in "AB" form, "a" is an integer without a leading 0 (greater than 0, not more than 1e9), and "B" is a number (possibly empty) factorial symbol ("!"). For example: 3!! =6!=720Solution: Set two large integer form a part of the A,b;b section respectively has n1,n2 a symbol. Assuming N1 > n2, then we just need to determine the size of AA and B, where A is (N1-N2) a factorial symbol. N1 = N2 or N1 My

[j]*x[K] + b[j]*y[K] + c)sQ (d)* (sQ (a[j])+sQ (b[j]))) ans2++; } ans = max (ans, ans2); } } for(inti =0; I x. Size (); i++) { for(intj = i +1; J x. Size (); J + +) {Double X1 = (x[I] +x[j]) *0.5, y1 = (y[I] +y[j]) *0.5; for(intK =0; K 4; k++) {Double c =-(a[k] * x1 + b[k] * y1);intAns2 =0; for(intL =0; L x. Size (); l++) {if(sQ (a[k]*x[l] + b[k]*y[l] + c)sQ (d)* (sQ (a[k])+sQ (b[k]))) ans2++; } ans = max (ans, ans2); } } }returnAns }}tt; Copyri

Returns: "1:1:1" 2) 5436 Returns: "1:30:36" 3)

ints R1, C1, R2, and C2. compute and return the smallest number of moves a bishop needs to get from (R1, C1) to (R2, C2 ). if it is impossible for a bishop to reach the target cell, return-1 instead.DefinitionClass:BishopmoveMethod:HowmanymovesParameters:Int, intReturns:IntMethod signature:Int howmanymoves (INT R1, int C1, int R2, int C2)(Be sure your method is public)LimitsTime limit (s ):2.000Memory limit (MB ):256Constraints-R1, C1, R2, C2 will be between 0 and 7, aggressive.Examples0) 4673R

"}"Abcdefghi"Returns: 1There is only one way to trace this path. Each letter is used exactly once.1) ????{"ABC ","Fed ","Gai "}"Abcdea"Returns: 2Once we get to the 'E', we can choose one of two directions ctions for the final 'A '.2) ????{"ABC ","Def ","Ghi "}"ABCD"Returns: 0We can trace a path for "ABC", but there's no way to complete a path to the letter 'D '.3) ????{"AA ","AA "}"Aaaa"Returns: 108We can start from any of the four locations. from each location, we can then move in any of the t

) "3" Returns: { "NONE", "NONE" } 5) "12221112222221112221111111112221111" Returns: { "01101001101101001101001001001101001", "10110010110110010110010010010110010" }

See: http://robotcator.logdown.com/posts/231132-topcoder-srm-623 We recommend that you use the plug-in greed 2.0, which is very useful. But I don't know how to add test data by myself. I will study it next time.Greed 2.0 https://github.com/shivawu/topcoder-greed250ptThere are n trees on the ring runway, marked as 1 -- N. Some trees are recorded when Alice is running. The number is used to calculate the min

Topcoder srm 623 solution report We recommend that you use the plug-in greed 2.0, which is very useful. But I don't know how to add test data by myself. I will study it next time.Greed 2.0 https://github.com/shivawu/topcoder-greed250ptThere are n trees on the ring runway, marked as 1 -- n. Some trees are recorded when Alice is running. The number is used to calculate the minimum number of laps.Question: If

I wrote a topcoder exercise question a few days ago. +Code+ CommentsArticleNow, I will write the second article in this series.Note: There will be three articles in this series, a complete set of SRM questions I have done before, with 250, 500, and 1000 points, this article will introduce the 500 points. afterwards, I will occasionally take the time to participate in the latest SRM, and then offer the questions and answers to them :) Source: srm250 d

Topcoder srm287 div2 report Match Date: Saturday, February 4, 2006REPORT Date: Thursday, February 09,200 6 Preface:This is the first time to participate in topcoder. Although the three questions were submitted before the conclusion, but the accuracy was very low, so not only gave another person a challenge, but also left only one question after system test. However, we are lucky because problem-600 is a tou

[ -][ -][ -][ -];intMain () {FO (RPS); N=gi; for(intI=0; i) {R[i]=gi; R[i]/= -; P[i]=gi; P[i]/= -; S[i]=gi; S[i]/= -; } for(intA=0; a) {dp[a][0][0][0]=1.0; for(intI=0; i) for(intj=i;j+1; j--) for(intk=i-j;k+1; k--) for(ints=i-j-k;s+1; s--) { if(i!=a) {Dp[a][j+1][k][s]=dp[a][j+1][k][s]+1.0* (j+k+s+1)/(i+1) *r[i]*Dp[a][j][k][s]; Dp[a][j][k+1][s]=dp[a][j][k+1][s]+1.0* (j+k+s+1)/(i+1) *p[i]*Dp[a][j][k][s]; Dp[a][j][k][s+1]=

)==0) ans =ABS(x) +ABS(y);ElseAns =ABS(x) +ABS(y) +2; }Else if(x 0 y 0) {if((x1)==1|| (y1)==1) ans =ABS(x) +ABS(y) +2;ElseAns =ABS(x) +ABS(y) +4; }Else if(X >0 y 0) {if((x1)==1|| (y1)==0) ans =ABS(x) +ABS(y);ElseAns =ABS(x) +ABS(y) +2; }Else if(X 0 y >0) {if((x1)==0|| (y1)==1) ans =ABS(x) +ABS(y);ElseAns =ABS(x) +ABS(y) +2; }returnAns }};The first method of code#include #include #include #include using namespace STD;structdata{intx, y;intIdintD;} d[100010];intnd queueQintTin

, found that there is a better wording, do not need to pre-processing what, direct O (n) sweep, with two variables sentenced to the line.Best Code#include using namespace STD;classniceorugly{ Public:BOOLIsvol (Charc) {returnc = =' A '|| c = =' E '|| c = =' I '|| c = =' O '|| c = =' U '; }stringDescribestrings) {BOOLCan1 =0;intNV, NC; NV = NC =0; for(inti =0; I 0; i++) {if(S[i] = ='? ') nv++, nc++;Else if(Isvol (S[i])) nv++, NC =0;ElseNV =0, nc++;if(NV >=3|| NC >=5) Can1 =1; }BOOLCan2 =1; NV = NC

{7} {2} Returns:2 There is three teams. Your team have 1 point and each of the other teams have 7 points. With three teams, the remaining matches is determined Uniquely:each pair of teams must play a single match against each Other. The best possible final result for your team was

Give you a 01 matrix, find the largest of the 01 alternating matricesBecause n Max is only 100, so direct violent promiscuityFirst, the line I, all columns up the legal length, and then enumerate the column J as the leftmost columns, calculate the maximum matrix can be obtained/************************************************************************* > File name:2.cpp > Author:al ex > Mail: [email protected] > Created time:2015 May 07 Thursday 15:07 58 seconds ***********************************