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=0;
    int flag2=0;
    int flag3=0;
    int flag4=0;
    while(_--){
        string s;
        cin>>s;
        if(s==str){
            flag4=1;
            continue;
        }
        if(s[0]==str[1]&&s[1]==str[0]){
            flag3=1;
            continue;
        }
        if(str[0]==s[1]){
            flag1=1;
        }
        if(str[1]==s[0]){
            flag2=1;
        }
    }
//    cout<<flag1<<" "<<flag2<<" "<<flag3<<" "<<flag4; 
    if(flag3||flag4||(flag1+flag2==2)){
        printf("YES
");
    }else{
        printf("NO
");
    }
    return 0; 
} 
/*
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[500];
signed main(){
    int h,m, s, t1, t2;
    cin>>h>>m>>s>>t1>>t2;
    h*=5;t1*=5;
    t2*=5;
    vis[h]++;
    vis[m]++;
    vis[s]++;
    int ss=0;
    if(t1>t2)
        swap(t1,t2);
    for(int i=t1;i<t2;i++){
        ss+=vis[i];
    }
    if(ss==0||ss==3){
        cout<<"YES";
    }
    else{
        cout<<"NO";
    }
    return 0;
}

 题解：二进制+暴力。

#include<bits/stdc++.h>
using namespace std;
int a[150000];
int vis[150];int n,k;
signed main(){
    memset(vis,0,sizeof(vis));
    cin>>n>>k;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=k;j++){
            scanf("%d",&a[j]);
        }
         vis[a[1]+a[2]*2+a[3]*4+a[4]*8]=1;
    }
    for(int i=0;i<=(1<<k);i++){
        for(int j=i;j<=(1<<k);j++){
            if(vis[i]&&vis[j]&&(!(i&j))){
                 printf("YES");    
                 return 0;
            }
        }
    }
    printf("NO");
    return 0;
}