SMU Summer 2023 Contest Round 4

SMU Summer 2023 Contest Round 4

A. Telephone Number

满足第一个8后面存在10个字符即可

#include  <bits/stdc++.h>
#define endl '\n'
#define int long long
 
using namespace std;
 
int n,m;
void solve(){
    cin >> n;
    string s;
    cin >> s;
    int p = s.find_first_of('8') ;
    if(p > n - 11 || p == -1 || n < 11)
        cout << "NO" << endl;
    else
        cout << "YES" << endl;
}
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);cout.tie(nullptr);
    int Ke_scholar = 1;
    cin >> Ke_scholar;
    while(Ke_scholar--)
        solve();
    return 0;
}
 

B. Lost Numbers

因为只有6个数,所以可以直接用\(next\_permutation\)全排列暴力匹配,不要关闭流同步!!!

#include  <bits/stdc++.h>
#define endl '\n'
#define int long long
 
using namespace std;
 
int n,m;
void solve(){
 
    int res[5];
    for(int i = 1;i <= 4;i ++){
        cout << "? " << i << ' ' << i + 1 << endl;
        fflush(stdout);
        cin >> n;
        res[i] = n;
    }
 
    int ans[] = {0,4,8,15,16,23,42};
    do{
        if(ans[1] * ans[2] == res[1] && ans[2] * ans[3] == res[2] && ans[3] * ans[4] == res[3] && ans[4] * ans[5] == res[4]){
            cout << '!';
            for(int i = 1;i <= 6;i ++)
                cout << ' ' << ans[i];
            cout << endl;
            fflush(stdout);
            break;
        }
    }while(next_permutation(ans + 1,ans + 7));
 
}
signed main()
{
//    ios::sync_with_stdio(false);
//    cin.tie(nullptr);cout.tie(nullptr);
    int Ke_scholar = 1;
//    cin >> Ke_scholar;
    while(Ke_scholar--)
        solve();
    return 0;
}

C. News Distribution

用并查集维护每个人所在的圈子,然后循环每个人所在的圈子\(++\)

#include  <bits/stdc++.h>
#define endl '\n'
#define int long long
 
using namespace std;
 
int n,m;
void solve(){
    cin >>n >> m;
 
    vector<int> fa(n + 1);
    vector<int> ans(n + 1,0);
    iota(fa.begin(),fa.end(),0);
 
    auto find = [&](auto self,int x)->int {
        return fa[x] == x ? x : (fa[x] = self(self,fa[x]));
    };
 
    auto add = [&](int x,int y){
        fa[find(find,x)] = find(find,y);
    };
 
    for(int i = 1;i <= m;i ++){
        int k,o,p;
        cin >> k;
        if(k) cin >> o;
        for(int j = 1;j < k;j++){
            cin >> p;
            add(o,p);
        }
    }
    for(int i = 1;i <= n;i++)
        ans[find(find,i)]++;
    for(int i = 1;i <= n;i ++){
        cout << ans[find(find,i)] << ' ';
    }
 
}
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);cout.tie(nullptr);
    int Ke_scholar = 1;
//    cin >> Ke_scholar;
    while(Ke_scholar--)
        solve();
    return 0;
}

D. Bicolored RBS

\(r\)和\(b\)左右括号每相隔一个交叉染色即可

#include  <bits/stdc++.h>
#define endl '\n'
#define int long long
#define  inf 0x3f3f3f3f
 
using namespace std;
 
int n,m;
int zr,zb;
void solve(){
    string s;
    cin >> n >> s;
    stack<char> r,b;
    for(int i = 0;i < n;i ++){
        if(s[i] == '('){
            if(zr & 1){
                cout << 0;
            }else{
                cout << 1;
            }
            zr++;
        }else{
            if(zb & 1){
                cout << 0;
            }else{
                cout << 1;
            }
            zb++;
        }
    }
}
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);cout.tie(nullptr);
    int Ke_scholar = 1;
//    cin >> Ke_scholar;
    while(Ke_scholar--)
        solve();
    return 0;
}