D:找到两个数 一个是另一个的整数倍(1也算)

因为N是600000 调和级数为ln(n+1) 算一下 可以直接爆

#include 
      
       #include 
       
        #include 
        
         #include 
         
          #define foror(i,a,b) for(i=a;i
          
           b;i--)#define EPS 1.0e-8#define PI acos(-1.0)#define INF 3000000000#define MOD 1000000009#define mem(a,b) memset((a),b,sizeof(a))#define TS printf("!!!\n")#define lson o<<1, l, m#define rson o<<1|1, m+1, r//using ll = long long;//using ull= unsigned long long;//std::ios::sync_with_stdio(false);using namespace std;//priority_queue
           
            
             ,greater
             
              > que;typedef long long ll;bool cmp(int x,int y){ return x>y;}int where[600005];int visit[600005];int main(){ //freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); int n,k; cin >> n >> k; int ans1=0,ans2=0; int t; for(int i=1;i<=k;i++) { scanf("%d",&t); if(visit[t]!=0) //==1 { ans1=where[t]; ans2=i; break; } visit[t]++; where[t]=i; } if(ans1!=0&&ans2!=0) { printf("%d %d\n",ans1,ans2); return 0; } for(int i=1;i<=n;i++) for(int j=2*i;j<=n;j+=i) { if(visit[i]!=0&&visit[j]!=0) { ans1=where[i]; ans2=where[j]; break; } } if(ans1!=0&&ans2!=0) { printf("%d %d\n",ans1,ans2); return 0; } printf("0 0\n"); return 0;}
             
            
           
          
         
        
       
      

E:两个板之间的数之和是K的倍数?? 反正这么做过了

#include 
      
       #include 
       
        #include 
        
         #include 
         
          #define foror(i,a,b) for(i=a;i
          
           b;i--)#define EPS 1.0e-8#define PI acos(-1.0)#define INF 3000000000#define MOD 1000000009#define mem(a,b) memset((a),b,sizeof(a))#define TS printf("!!!\n")#define lson o<<1, l, m#define rson o<<1|1, m+1, r//using ll = long long;//using ull= unsigned long long;//std::ios::sync_with_stdio(false);using namespace std;//priority_queue
           
            
             ,greater
             
              > que;typedef long long ll;bool cmp(int x,int y){ return x>y;}char a[100005];//(a1+an)*n/2;int main(){ //freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); int n,k; cin >> n; scanf("%s",a+1); //printf("%s\n",a+1); int num=0; int len=strlen(a+1); //cout <
              
               <
              
             
            
           
          
         
        
       
      

G:水

H:2*3的砖铺 N*M的地板 以前CF好像做过 直接判min(n,m)>=2&&max(n,m)>=3  然后n*m%6==0

证明的话 引用别人的:

乘积是6的倍数那就说明有因子2和3：

1)一边是2的倍数恰好另一边是3的倍数，这种情况下显然可行；

2)一边是6的倍数另一边不是2也不是3的倍数，那另一边肯定是奇数，为1的时候不行，设边为N=2a+1=2(a-1)+3，那前面就以2x3的形式放最后一个以3x2的形式放，由于一边为6的倍数那么就是一定可以放得下的

#include 
      
       #include 
       
        #include 
        
         #include 
         
          #define foror(i,a,b) for(i=a;i
          
           b;i--)#define EPS 1.0e-8#define PI acos(-1.0)#define INF 3000000000#define MOD 1000000009#define mem(a,b) memset((a),b,sizeof(a))#define TS printf("!!!\n")#define lson o<<1, l, m#define rson o<<1|1, m+1, r//using ll = long long;//using ull= unsigned long long;//std::ios::sync_with_stdio(false);using namespace std;//priority_queue
           
            
             ,greater
             
              > que;typedef long long ll;int main(){ //freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); int k; cin >> k; while(k--) { ll n,m; cin >> n >> m; ll sum=n*m; if(min(n,m)<2||max(n,m)<3) { cout<<"No"<
             
            
           
          
         
        
       
      

I:阅读理解 把最长的R B先连起来

#include 
      
       #include 
       
        #include 
        
         #include 
         
          #define foror(i,a,b) for(i=a;i
          
           b;i--)#define EPS 1.0e-8#define PI acos(-1.0)#define INF 3000000000#define MOD 1000000009#define mem(a,b) memset((a),b,sizeof(a))#define TS printf("!!!\n")#define lson o<<1, l, m#define rson o<<1|1, m+1, r//using ll = long long;//using ull= unsigned long long;//std::ios::sync_with_stdio(false);using namespace std;//priority_queue
           
            
             ,greater
             
              > que;typedef long long ll;int red[1005],blue[1005];bool cmp(int x,int y){ return x>y;}int main(){ //freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); int n; while(scanf("%d",&n)==1) { mem(red,0); mem(blue,0); int pop=0; int pop1=0; int number=0; int ans=0; char ch; int j; for(int i=1;i<=n;i++) { cin >> j>>ch; if(ch=='R') { red[++pop]=j; } else { blue[++pop1]=j; } } sort(red+1,red+pop+1,cmp); sort(blue+1,blue+pop1+1,cmp); for(int i=1;i<=min(pop,pop1);i++) { ans+=red[i]+blue[i]-2; } cout <
              
               <
              
             
            
           
          
         
        
       
      

J:按条件读取字符 组成词组 输出

#include 
      
       #include 
       
        #include 
        
         #include 
         
          #define foror(i,a,b) for(i=a;i
          
           b;i--)#define EPS 1.0e-8#define PI acos(-1.0)#define INF 3000000000#define MOD 1000000009#define mem(a,b) memset((a),b,sizeof(a))#define TS printf("!!!\n")#define lson o<<1, l, m#define rson o<<1|1, m+1, r//using ll = long long;//using ull= unsigned long long;//std::ios::sync_with_stdio(false);using namespace std;//priority_queue
           
            
             ,greater
             
              > que;typedef long long ll;//stack
              
                a;int pop=0;map
               
                 mp;string strans;int ans=0;char c[550];//(a1+an)*n/2;void print(){for(map
                
                 ::iterator it=mp.begin();it!=mp.end();it++){ if(it->second>ans) { strans=it->first; ans=it->second; }}cout << strans<<":"<
                 
                  <
                  
                   ='a'&&c[i]<='z') { int cur=i; while(c[cur]>='a'&&c[cur]<='z') { start+=c[cur]; cur++; } i=cur; if(c[cur]==','||c[cur]=='\0') continue; start+=' '; while(c[cur]==' ') cur++; while(c[cur]>='a'&&c[cur]<='z') { start+=c[cur]; cur++; } //cout <<"start="<
                   
                    <<" "; //cout <