A - Sequence of Strings

#include<bits/stdc++.h>
using namespace std;

int32_t main(){
    int n;
    cin >> n;
    vector<string> s(n);
    for( auto &i : s ) cin >> i;
    for( int i = n-1 ; i >= 0 ; i -- )
        cout << s[i] << "\n";
    return 0;
}

B - Multi Test Cases

#include<bits/stdc++.h>
using namespace std;

int read(){...}

void solve(){
    int n = read() , res = 0;
    for( int x ; n ; n -- ){
        x = read();
        if( x & 1 ) res ++;
    }
    cout << res << "\n";
    return;
}

int32_t main(){
    for( int T = read() ; T ; T -- )
        solve();
    return 0;
}

C - Count Connected Components

#include<bits/stdc++.h>
using namespace std;

int read(){...}

vector<int> vis;
vector<vector<int>> e;

void dfs( int u ){
    vis[u] = 1;
    for( auto v : e[u] ){
        if( vis[v] ) continue;
        dfs(v);
    }
    return;
}

int32_t main(){
    int n = read() , m = read() , res = 0;
    e.resize(n+1) , vis.resize(n+1);

    for( int u , v ; m ; m -- )
        u = read() , v = read() , e[u].push_back(v) , e[v].push_back(u);
    for( int i = 1 ; i <= n ; i ++ ){
        if( vis[i] ) continue;
        dfs(i) , res ++;
    }
    cout << res << "\n";
    return 0;
}

D - Happy New Year 2023(补题)

根据唯一分解定理，\(p^2q\)是唯一的，且\(max(p,q)\le \sqrt[3]{n}\)

所以直接暴力写就好

#include<bits/stdc++.h>
#define int unsigned long long
typedef long long ll;
using namespace std;

int read(){...}

void solve(){
    int n = read();
    for( int i = 2 , j ; ; i ++ ){
        if( n % i != 0 ) continue;
        n /= i;
        if( n % i == 0 ) j = n / i;
        else j = i , i = sqrt( n );
        cout << i << " " << j << "\n";
        return;
    }
}

int32_t main(){
    for( int T = read() ; T ; T -- ) solve();
    return 0;
}

E - Count Simple Paths

非常简单的统计一下从1开始的简单路径，dfs不断的搜索就好了。

#include<bits/stdc++.h>
using namespace std;

int read(){...}

const int N = 1e6 ;
int n , m , res;
vector<int> vis;
vector<vector<int>> e;

void dfs( int u ){
    vis[u] = 1;
    res ++;
    if( res == N ){
        cout << res<<"\n";
        exit(0);
    }
    for( auto v : e[u] ){
        if( vis[v] ) continue;
        dfs(v);
    }
    vis[u] = 0;
    return;
}


int32_t main(){
    n = read() , m = read() , res = 0;
    e.resize(n+1) , vis.resize(n+1);
    for( int u , v ; m ; m -- ){
        u = read() , v = read();
        e[u].push_back(v) , e[v].push_back(u);
    }
    dfs(1);
    cout << res << "\n";
    return 0;
}

F - ABCBAC（补题）