Codeforces Round #438 by Sberbank and Barcelona Bootcamp (Div. 1 + Div. 2 combined) A,B,C【真的菜·】

8说了

 #include<bits/stdc++.h>

 using namespace std;
 #define int long long

 signed main(){
     string str;
     cin>>str;
     int _;
     cin>>_;
     int flag1=;
     int flag2=;
     int flag3=;
     int flag4=;
     while(_--){
         string s;
         cin>>s;
         if(s==str){
             flag4=;
             continue;
         }
         if(s[]==str[]&&s[]==str[]){
             flag3=;
             continue;
         }
         if(str[]==s[]){
             flag1=;
         }
         if(str[]==s[]){
             flag2=;
         }
     }
 //    cout<<flag1<<" "<<flag2<<" "<<flag3<<" "<<flag4;
     if(flag3||flag4||(flag1+flag2==)){
         printf("YES\n");
     }else{
         printf("NO\n");
     }
     return ;
 }
 /*
 ah
 oy

 aa

 ha
 yo
 */

思路：直接暴力---居然能过.钟表问题转换是真滴难QAQ【小学就没整懂】

 #include<bits/stdc++.h>

 using namespace std;
 #define int long long
 int vis[];
 signed main(){
     int h,m, s, t1, t2;
     cin>>h>>m>>s>>t1>>t2;
     h*=;t1*=;
     t2*=;
     vis[h]++;
     vis[m]++;
     vis[s]++;
     int ss=;
     if(t1>t2)
         swap(t1,t2);
     for(int i=t1;i<t2;i++){
         ss+=vis[i];
     }
     if(ss==||ss==){
         cout<<"YES";
     }
     else{
         cout<<"NO";
     }
     return ;
 }

题解：二进制+暴力。

 #include<bits/stdc++.h>
 using namespace std;
 int a[];
 int vis[];int n,k;
 signed main(){
     memset(vis,,sizeof(vis));
     cin>>n>>k;
     for(int i=;i<=n;i++){
         for(int j=;j<=k;j++){
             scanf("%d",&a[j]);
         }
          vis[a[]+a[]*+a[]*+a[]*]=;
     }
     for(int i=;i<=(<<k);i++){
         for(int j=i;j<=(<<k);j++){
             if(vis[i]&&vis[j]&&(!(i&j))){
                  printf("YES");
                  return ;
             }
         }
     }
     printf("NO");
     return ;
 }