BestCoder Round #41 -- (A, B)

BestCoder Round #41 -- (A, B)

 

Question transfer: BestCoder Round #41

 

 

A. ZCC loves straight flush

 

Train of Thought: simple question, but I didn't see the question at the beginning. wa had done it several times before I found that the input was not continuous. That is to say, each card of the same color can be placed in one piece, I don't need to care about the input order. I feel it should be clear about the question here.

 

AC code ):

 

#include 
 
  #include 
  
   #include 
   
    #include #include 
    
     #include 
     
      #include 
      
       #include 
       
        #include
        #include 
         
          #include 
          
           #include 
           
            #define LL long long#define INF 0x7fffffffusing namespace std;char a[5][5];int fun(int *p, int n) { sort(p, p + n); int ret = 1; if(p[0] != 1) { for(int i = 1; i <= 9; i ++) { int cur = 0; for(int j = 0; j < n; j ++) { if(p[j] <= i + 4 && p[j] >= i) { cur ++; } ret = max(ret, cur); } } } else if(p[0] == 1) { for(int i = 1; i <= 10; i ++) { int cur = 0; for(int j = 0; j < n; j ++) { if(p[j] <= i + 4 && p[j] >= i) { cur ++; } } if(i == 10) cur ++; ret = max(ret, cur); } } return ret;}int main() { int T; scanf("%d", &T); while(T --) { scanf("%s %s %s %s %s", a[0], a[1], a[2], a[3], a[4]); int cur = 1; int num[5]; memset(num, 0, sizeof(num)); for(int i = 0; i < 5; i ++) { int len = strlen(a[i]); for(int j = 1; j < len; j ++) { num[i] = num[i] * 10 + a[i][j] - '0'; } } /*for(int i = 0; i < 5; i ++) { cout << num[i] << " "; } cout << endl;*/ int A[5], B[5], C[5], D[5]; int an = 0, bn = 0, cn = 0, dn = 0; for(int i = 0; i < 5; i++) { if(a[i][0] == 'A') A[an ++] = num[i]; else if(a[i][0] == 'B') B[bn ++] = num[i]; else if(a[i][0] == 'C') C[cn ++] = num[i]; else if(a[i][0] == 'D') D[dn ++] = num[i]; } int ans = 1; ans = max(ans, fun(A, an)); ans = max(ans, fun(B, bn)); ans = max(ans, fun(C, cn)); ans = max(ans, fun(D, dn)); printf("%d\n", 5 - ans); } return 0;}
           
          
         
       
      
     
    
   
  
 

 

 

 

 

B. ZCC loves strings

 

Train of Thought: Find the rule, simple game, but there are many pitfalls, there are a total of Cn in 2, and then Miss G wins an odd and even number together, when selecting the same string, determine whether the ans is 0 and then obtain a gcd. Remember to use longlong to blow up the int.

 

AC code:

 

#include 
 
  #include 
  
   #include 
   
    #include #include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         #include
         #include 
          
           #include 
           
            #include 
            
             #define LL long long#define INF 0x7fffffffusing namespace std;int a[200005];LL gcd(LL a, LL b) { return b == 0 ? a : gcd(b, a % b);}int main() { ios::sync_with_stdio(false); int T, n; scanf("%d", &T); while(T --) { scanf("%d", &n); string tmp; LL ji = 0, ou = 0; map
             
               mmp; LL jia = 0; for(int i = 0; i < n; i ++) { cin >> tmp; if(mmp.find(tmp) != mmp.end()) { jia += mmp[tmp]; mmp[tmp] ++; } else mmp[tmp] = 1; a[i] = tmp.length(); if(a[i] & 1) ji ++; else ou ++; } LL ans = ji * ou; LL sum = n * (n - 1) / 2; ans += jia; if(ans == 0) { printf("0/1\n"); continue; } LL gg = gcd(ans, sum); ans /= gg; sum /= gg; cout << ans << '/' << sum << endl; } return 0;}
             
            
           
          
        
       
      
     
    
   
  
 

 

 

 

 

 

Okay, this is a little better game I 've been playing with algorithms for more than half a year. Although I 've been playing a game for the last few minutes, I'm still very excited about the hack process. It's very touching, BC purple name, although the quality is not too high, but very excited, quietly waiting for my CF purple name, first posted an image to commemorate