Codeforces Round #561 (Div. 2) ABC题解-白红宇

Codeforces Round #561 (Div. 2) ABC题解

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发布时间：2019-05-23

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虽说是退役了，但是上分这种事还是放不下滴，有空打打比赛，锻炼下智商

题目链接：

A. Silent Classroom

统计相同首字母的单词数目，然后每个字母群的最小分组为num*(num-1)/2;

参考代码：

//include 
   
    #include 
    
     #include 
     #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             #include 
             
              #include 
              
               #include 
               
                #include 
                
                 #include 
                 
                  using namespace std;typedef long long ll;typedef unsigned long long ull;const ll MOD = 10007;const int maxn = 2*1e5+5;const int INF = 0x3f3f3f3f;const ll LINF = 0x3f3f3f3f3f3f3f3f;template 
                  
                   inline bool scan_d(T &ret) { char c; int sgn; if (c = getchar(), c == EOF) return 0; while (c != '-' && (c<'0' || c>'9')) c = getchar(); sgn = (c == '-') ? -1 : 1; ret = (c == '-') ? 0 : (c - '0'); while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0'); ret *= sgn; return 1;}int n;int num[30];int main(){ memset (num,0,sizeof(num)); scanf("%d",&n); for (int i=0;i
                   
                    0) { int t=num[i]/2; sum =sum+t*(t-1)/2; t=num[i]-t; sum =sum+t*(t-1)/2; } } printf("%lld\n",sum); return 0;}

B. All the Vowels Please

给出的数k只要分解的n,m都大于5,则就可以组成，答案就是排出5*5矩阵之后每一行多出的随便补就行了。

代码如下：

//include 
   
    #include 
    
     #include 
     #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             #include 
             
              #include 
              
               #include 
               
                #include 
                
                 #include 
                 
                  using namespace std;typedef long long ll;typedef unsigned long long ull;const ll MOD = 10007;const int maxn = 2*1e5+5;const int INF = 0x3f3f3f3f;const ll LINF = 0x3f3f3f3f3f3f3f3f;template 
                  
                   inline bool scan_d(T &ret) { char c; int sgn; if (c = getchar(), c == EOF) return 0; while (c != '-' && (c<'0' || c>'9')) c = getchar(); sgn = (c == '-') ? -1 : 1; ret = (c == '-') ? 0 : (c - '0'); while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0'); ret *= sgn; return 1;}int k;char a[6][10]={"aeiou","eioua","iouae","ouaei","uaeio"};int main(){ //a[0]="aeiou"; //a[1]="eioua"; //a[2]="iouae"; //a[3]="ouaei"; //a[4]="uaeio"; scanf("%d",&k); int n=0,m=0,flag=0; for (int i=5;i<=k;i++) { if(k%i==0) { n=i; m=k/i; if(m>=5) { flag=1; break; } } } if(flag==0) { printf("-1\n"); } else { for (int i=0;i

C. A Tale of Two Lands

打表发现了规律，固定X，只要是X/2+X%2=<Y<=2*X(可互换位置)，-（X/2+X%2）=<Y<=-（2*X）（可互换位置）,二分查找上界和下界即可。

代码如下：

//include 
   
    #include 
    
     #include 
     #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             #include 
             
              #include 
              
               #include 
               
                #include 
                
                 #include 
                 
                  using namespace std;typedef long long ll;typedef unsigned long long ull;const ll MOD = 10007;const int maxn = 2*1e5+5;const int INF = 0x3f3f3f3f;const ll LINF = 0x3f3f3f3f3f3f3f3f;template 
                  
                   inline bool scan_d(T &ret) { char c; int sgn; if (c = getchar(), c == EOF) return 0; while (c != '-' && (c<'0' || c>'9')) c = getchar(); sgn = (c == '-') ? -1 : 1; ret = (c == '-') ? 0 : (c - '0'); while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0'); ret *= sgn; return 1;}int n;ll a[maxn];vector
                   
                    z,f;int num=0;int main(){ scanf("%d",&n); for (int i=1;i<=n;i++) { scanf("%lld",&a[i]); if(a[i]>0) { z.push_back(a[i]); } else if(a[i]<0) { f.push_back(-a[i]); } else { num++; } } ll sum=0; sort(z.begin(),z.end()); sort(f.begin(),f.end()); for (int i=1;i<=n;i++) { if(a[i]!=0) { if(a[i]<0) a[i]=-a[i]; int loca,locb; loca=upper_bound(z.begin(),z.end(),a[i]*2)-z.begin(); locb=lower_bound(z.begin(),z.end(),a[i]/2+a[i]%2)-z.begin(); sum=sum+loca-locb; loca=upper_bound(f.begin(),f.end(),a[i]*2)-f.begin(); locb=lower_bound(f.begin(),f.end(),a[i]/2+a[i]%2)-f.begin(); sum=sum+loca-locb; sum--; } } sum/=2; sum =sum+(num)*(num-1)/2; printf("%lld\n",sum); return 0;}