Codeforces #200 (div.2) simulated exercise Competition

Codeforces #200 (div.2) simulated exercise Competition
Question A: How many ideas can A magnet be divided into when the same-sex magnet excludes each other? Read and calculate while reading and adding 1 when reading something different from the previous one (the first group of data is specially judged) I did not clarify my thinking and try again, but I was too full of = _ + than my teammates. Code:

#include 
  
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            #include using namespace std;typedef long long int LL;const int M = 100009,INF = 0x3fffffff;int main(void) { //problem: cf#200, address: int n, y, last = -1, ans = 1; string x; cin >> n; while(n--) { cin >> x; if(x == "10") y = 10; else y = 1; if(last != -1 && y != last) ans++; last = y; } cout << ans << endl; return 0;}
           
          
         
        
       
      
     
    
  

Question B: given three atomic numbers (A, B, C), each atom has A chemical price, indicating the number of chemical bonds to be connected to other atoms (however, the figure is not good, misleading A lot of time) solution: as long as you know the number of chemical bonds between A and B atoms X , The number of atoms of the other two edges can contain X And then check whether it meets the atomic price requirements. Code:

#include 
  
   #include
   #include 
    
     #include 
     
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       #include 
       
        #include 
        
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          #include 
          
           #include 
           
            #include using namespace std;typedef long long int LL;const int M = 100009,INF = 0x3fffffff;int main(void) { //problem:cf#200 , address: int a, b, c; cin >> a >> b >> c; bool ok = false; for(int i = 0; i <= b; i++) { if(b - i == c - (a - i)) { ok = true; cout << i << " " << b - i << " "<< a - i << endl; break; } } if(!ok) cout << "Impossible" << endl; return 0;}
           
          
         
        
       
      
     
    
  

Question C: given an unlimited unit resistance, each operation can give the current circuit parallel or concatenate a unit resistance to reach the target resistance (in the form of a false score) the minimum number of operations required for resistance. Idea: the idea is to start from a unit of resistance and search for the minimum number of Target solutions with breadth first, however, the exponential size of the forward solution space of 2 must be a queue or time-out. Then we decided to roll back from the answer to the initial state, obviously reducing the space size. Sometimes it is a good idea to start from the target solution, which can greatly reduce the solution space. This idea was used in question A of the weekly competition in the afternoon and was successfully implemented. Note: In this question, the long data is used. However, the long data used for calculation in the middle of the two data I read is still int and WA. Code:

#include 
  
   #include
   #include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include using namespace std;typedef long long int LL;const int M = 100009,INF = 0x3fffffff;LL a, b, ans;void dfs(LL x, LL y) { if(y == 0) return; ans += x / y; x %= y; swap(x, y); dfs(x, y);}int main(void) { //problem: cf#200, address: cin >> a >> b; dfs(a, b); cout << ans << endl; return 0;}
           
          
         
        
       
      
     
    
  

D: Question: Can I solve the Winding Rope? Idea: as long as the two connected overwrites are the same line, the two overwrites can be directly eliminated. In this way, it is easy to use a stack to read and simulate. Code:

#include 
  
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           #include 
           
            #include using namespace std;typedef long long int LL;const int M = 100009,INF = 0x3fffffff;int main(void) { //problem: , address: char c; stack
            
              s; while ((c = getchar()) != '\n') { int x = (c == '+' ? 1 : -1); if(s.empty()) {s.push(x); continue;} if(x == s.top()) s.pop(); else s.push(x); } cout << (s.empty() ? "Yes" : "No") << endl; return 0;}
            
           
          
         
        
       
      
     
    
  

Conclusion: grasp code details and train thinking skills.