Codeforces Beta Round #61 (Div. 2)

A

输入用long double

1 #include
      
        2 using namespace std; 3 #define lson l,mid,rt<<1 4 #define rson mid+1,r,rt<<1|1 5 #define sqr(x) ((x)*(x)) 6 #define pb push_back 7 #define eb emplace_back 8 #define maxn 1000006 9 #define rep(k,i,j) for(int k=i;k
       
        >n;20     if(n>=-128&&n<=127) cout<<"byte"<
        
         =-32768&&n<=32767) cout<<"short"<
         
          =-2147483648&&n<=2147483647) cout<<"int"<
         
        
       
      

 

B

暴力枚举每一个数即可

1 #include
      
        2 using namespace std; 3 #define lson l,mid,rt<<1 4 #define rson mid+1,r,rt<<1|1 5 #define sqr(x) ((x)*(x)) 6 #define pb push_back 7 #define eb emplace_back 8 #define maxn 1000006 9 #define rep(k,i,j) for(int k=i;k
       
        >n;22     for(int i=1;i<=n;i++){23         cin>>a[i];24     }25     int ans=1;26     for(int i=1;i<=n;i++){27         int j=i-1;28         int co=1;29         while(j>=1){30             if(a[j+1]>=a[j]){31                 j--;32                 co++;33             }34             else{35                 break;36             }37         }38         j=i+1;39         while(j<=n){40             if(a[j-1]>=a[j]){41                 j++;42                 co++;43             }44             else{45                 break;46             }47         }48         if(co>ans) ans=co;49     }50     cout<
        
         <
        
       
      

 

C

模拟题

1 #include
      
        2 using namespace std; 3 #define lson l,mid,rt<<1 4 #define rson mid+1,r,rt<<1|1 5 #define sqr(x) ((x)*(x)) 6 #define pb push_back 7 #define eb emplace_back 8 #define maxn 1000006 9 #define rep(k,i,j) for(int k=i;k
       
         M,N;16 17 int main(){18     #ifndef ONLINE_JUDGE19         freopen("input.txt","r",stdin);20     #endif21     std::ios::sync_with_stdio(false);22     while(cin>>t)23     {24         s=t;25         int F=0;26         while(1)27         {28             int x=s.find_last_of('\\');29             if (x==2) break;30             s=s.substr(0,x);31             int f=N[s];32             M[s]+=F;33             N[s]++;34             res1=max(res1,M[s]);35             res2=max(res2,N[s]);36             if (!f) F++;37         }38     }39     cout<
        
         <<' '<
         
          <
         
        
       
      

 

D

找出3个数，使他们两两的公约数互不为1，他们三个的公约数为1，剩下的数就输出他们的倍数即可

1 #include
      
        2 using namespace std; 3 #define lson l,mid,rt<<1 4 #define rson mid+1,r,rt<<1|1 5 #define sqr(x) ((x)*(x)) 6 #define pb push_back 7 #define eb emplace_back 8 #define maxn 1000006 9 #define rep(k,i,j) for(int k=i;k
       
        >n;22     if(n==2){23         cout<<-1<
       
      

 

E

找出a[i]-b[i]前缀和的最小值，然后依次减去，如果发现minn大于等于0的情况，说明走的通，逆向同理

1 #include
      
        2 using namespace std; 3 #define lson l,mid,rt<<1 4 #define rson mid+1,r,rt<<1|1 5 #define sqr(x) ((x)*(x)) 6 #define pb push_back 7 #define eb emplace_back 8 #define maxn 1000006 9 #define rep(k,i,j) for(int k=i;k
       

se;17 set

::iterator it;18 19 void func(int x){20 int minn=a[0]-b[0],pre=minn;21 for(int i=1;i

=0){27 if(x){28 se.insert(i+1);29 }30 else{31 se.insert(n-i);32 }33 }34 minn-=a[i]-b[i];35 }36 }37 38 int main(){39 #ifndef ONLINE_JUDGE40 // freopen("input.txt","r",stdin);41 #endif42 std::ios::sync_with_stdio(false);43 cin>>n;44 rep(i,0,n) cin>>a[i];45 rep(i,0,n) cin>>b[i];46 func(1);47 reverse(a,a+n);48 reverse(b,b+n-1);49 func(0);50 cout<

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