SRM 513 div2

250pt, a big question, just simulate it according to the meaning of the question.

 
# Include <iostream> # include <cstdio> # include <string. h >#include <vector >#include <cmath> # include <algorithm> using namespace STD; Class trainingcamp {public: vector <string> determinesolvers (vector <string> attendance, vector <string> problemtopics) ;}; vector <string> trainingcamp: determinesolvers (vector <string> A, vector <string> P) {int n =. size (); int M = A [0]. size (); int K = P. size (); vector <string> B; B. clear (); string ant; ant. clear (); int I = 0, j = 0, T = 0; for (I = 0; I <n; I ++) {ant. clear (); For (j = 0; j <K; j ++) {for (t = 0; t <m; t ++) {If (P [J] [T] = '-' | (P [J] [T] = 'X' & A [I] [T] = = 'X ')) continue; If (P [J] [T] = 'X' & A [I] [T] = '-') break;} If (T <m) ant. push_back ('-'); else ant. push_back ('x');} B. push_back (ANT);} return B ;}
 

40000 PT, brute force, complexity O (* 50 ).

In the system test, a condition for determining the left vertex is missing.

While (L <= R-l [I] & L> B [pur])
PUR = pur + 1;

=!

#include 
  
   #include 
   
    #include 
    
     #include 
     
      #include 
      
        # include 
       
         using namespace STD; Class yetanotherincrediblemachine {public: int countways (vector 
        
          platformmount, vector 
         
           platformlength, vector 
          
            bils) ;}; int yetanotherincrediblemachine: countways (vector 
           
             M, vector 
            
              L, vector 
             
               B) {int lenm = m. size (); B. push_back (1000000); long ant = 1; long mod = 1000000009; sort (B. begin (), B. end (); int pur = 0; For (INT I = 0; I 
              
                B [pur]) PUR = pur + 1; for (Int J = L; j <= R-l [I]; j ++) {If (j = B [pur]) PUR = pur + 1; else if (J + L [I]