Codeforces beta round #2 exercises

Question A: Simulate the question. Just be careful.

Question B: DP

To give you a digital matrix, we need to go from the upper left corner to a path in the lower right corner. The number on this path is multiplied and the number at the end is the least.

The 0 at the end is produced by 2 and 5, so is it OK to think of the path with the least 2 or the path with the least 5?

Well, it's OK .*_*

 
Assume that the minimum number of numbers that walk from the upper left corner to the lower right corner of the page containing factor 2 is X, and the minimum number of five is Y, then the answer is Min (x, y ), that is to say, the number of the optimal solutions X and Y is the least proof: using the reverse proof method, we assume that the number of the optimal solutions contains a 2, B 5, a> X, B> y, the easy-to-obtain answer must be greater than min (x, y). Therefore, X and Y in the optimal solution must be the smallest.
 

View code

# Include <cstdio> # Include <Cmath> # Include < String > # Include <Cstdlib> # Include <Vector># Include <Map> # Include < Set > # Include <Iostream> # Include <Cstring> # Include <Algorithm> Using   Namespace  STD;  Const   Int INF = ~ 0u > 2  ; Const   Int M = 1010  ;  Int DP [m] [m] [ 2  ];  Int Min ( Int A, Int  B ){  Return A <B? A: B ;}  Int  Main (){  Int N, I, J, K, X, Y, Pos;  While (Scanf ( "  % D  " , & N )! = EOF) {pos =- 1  ; Memset (DP,  0 , Sizeof  (DP ));  For (I = 1 ; I <= N; I ++ ){ For (J = 1 ; J <= N; j ++ ) {Scanf (  "  % D  " , & X); y = X;  If (X = 0  ) {Pos = I;  Continue  ;}  While (Y % 2 = 0 ) {DP [I] [J] [ 0 ] ++; Y/= 2  ;}  While (X % 5 = 0 ) {DP [I] [J] [ 1 ] ++, X/= 5  ;}}}  For (I = 2 ; I <= N; I ++) {DP [I] [  1 ] [ 0 ] + = DP [I- 1 ] [ 1 ] [ 0  ]; DP [I] [  1 ] [ 1 ] + = DP [I- 1 ] [ 1 ] [ 1  ]; DP [  1 ] [I] [0 ] + = DP [ 1 ] [I- 1 ] [ 0  ]; DP [  1 ] [I] [ 1 ] + = DP [ 1 ] [I- 1 ] [ 1  ];}  For (I = 2 ; I <= N; I ++ ){ For (J = 2 ; J <= N; j ++ ) {DP [I] [J] [  0 ] + = Min (DP [I- 1 ] [J] [ 0 ], DP [I] [J- 1 ] [ 0  ]); DP [I] [J] [  1 ] + = Min (DP [I- 1 ] [J] [ 1 ], DP [I] [J- 1 ] [1  ]) ;}}  //  Printf ("% d \ n", DP [N] [N] [1], DP [N] [N] [0]);          String  Ans; k = DP [N] [N] [ 0 ]> DP [N] [N] [ 1  ];  If (Pos! =- 1 && DP [N] [N] [k]) {printf (  "  1 \ n "  );  For (I = 1 ; I <Pos; I ++) printf ( "  D  "  );  For (I = 1 ; I <n; I ++) printf ( "  R  "  );  For (I = Pos; I <n; I ++) printf ("  D  "  );  Continue  ;} I = N; j = N;  While ( 1  ){  If (DP [I- 1 ] [J] [k] <DP [I] [J- 1  ] [K]) {I --; Ans + = "  D  "  ;}  Else  {J -- ; Ans + = "  R  "  ;}  If (I = 1 ){  For (I = 1 ; I <j; I ++) ans + = "  R  "  ;  Break  ;}  If (J = 1  ){  For (J = 1 ; J <I; j ++) ans + ="  D  "  ;  Break  ;}}  Int Len = Ans. Length (); printf (  "  % D \ n  "  , DP [N] [N] [k]);  For (I = len- 1 ; I> = 0 ; I --) {Cout < Ans [I];} cout < Endl ;}  Return   0  ;} 

Question C: I will give you three circles, and find that the angle of a circle is equal to the angle of the three circles. If there are multiple circles, select the biggest point of the angle.

I used the climbing algorithm (Simulated Annealing), but it took a long time to adjust the parameters.

The evaluation function is the variance between the current point and the three circles. If a better solution is not obtained each time, the step is gradually reduced. If a better solution is obtained, the moving is always allowed.

When I go, I go up, down, left, right, and left. Maybe this is the reason why the answer is not accurate enough.

View code

# Include <cstdio> # Include <Cmath> Const   Double EPS = 1E- 6  ;  Int Dir [ 4 ] [ 2 ] = { 1 , 0 , 0 , 1 ,- 1 , 0 , 0 ,- 1  };  Struct Point {  Double  X, Y, R;} p [  10  ];  Int SGN ( Double  X ){  Return FABS (x) <EPS? 0 :( X> 0 ? 1 :- 1  );}  Double Dist (point a, point B ){  Return SQRT (A. x-b.x) * (A. x-b.x) + (A. y-b.y) * (A. Y- B. Y ));}  Double  Judge (point TMP ){  Int  I;  Double Sum = 0  ;  Double Ang [ 5  ];  For (I = 0 ; I < 3 ; I ++ ) {Ang [I] = Dist (P [I], TMP )/ P [I]. R; sum + = Ang [I];} sum /= 3  ;  Double Vir = 0  ;  For (I = 0 ; I < 3 ; I ++ ) Vir + = (ANG [I]-sum) * (ANG [I]-Sum );  Return  Vir ;}  Int  Main (){  Int  I, J, K;  Double X = 0 , Y = 0  ;  For (I = 0 ; I < 3 ; I ++ ) {Scanf (  " % Lf  " , & P [I]. X, & P [I]. Y ,& P [I]. R); x + = P [I]. X; y + = P [I]. Y;} X /= 3  ; Y /= 3  ; Point S, T; S. x = X; S. Y = Y;  Double Delta = 1  , Vir1, vir2;  While (Delta> EPS) {vir1 = Judge (s );  For (I = 0 ; I < 4 ; I ++ ) {T. X = S. x + Delta * dir [I] [ 0  ]; T. Y = S. Y + Delta * dir [I] [ 1  ]; Vir2 = Judge (t );  If (Vir2 <vir1)Break  ;}  If (I = 4 ) Delta * = 0.82 ; //  From 0.88 to 0.82 to get rid of the shadow of TLE.             Else  {S. x + = Delta * dir [I] [ 0  ]; S. Y + = Delta * dir [I] [ 1  ] ;}}  If (Judge (s) <EPS) printf (  "  %. 8lf %. 8lf \ n  "  , S. X, S. Y );  Return   0  ;}