A - Order Something Else

#include <bits/stdc++.h>

using namespace std;

#define int long long


int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
    int n , p , q;
    cin >> n >> p >> q;
    int res = p;
    for( int x ; n ; n -- ){
        cin >> x , res = min( res , q + x );
    }
    cout << res << "\n";
    return 0;
}

B - Strictly Superior

#include <bits/stdc++.h>

using namespace std;

#define int long long


int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
    int n, m;
    cin >> n >> m;
    vector<int> p(n + 1);
    vector<vector<int>> f(n + 1, vector<int>(m + 1));
    for (int i = 1, c; i <= n; i++) {
        cin >> p[i] >> c;
        for (int x; c; c--)
            cin >> x, f[i][x] = 1;
    }
    for( int i = 1 ; i <= n ; i ++ ){
        for( int j = 1 ; j <= n ; j ++ ){
            if( p[i] < p[j] ) continue;
            int q = 1 , t = 0 ;
            for( int l = 1 ; q && l <= m ; l ++ ){
                if( f[j][l] > f[i][l] ) t ++;
                else if( f[j][l] < f[i][l] ) q = 0;
            }
            if( q == 0 ) continue;
            if( p[i] > p[j] || t > 0 ){
                cout << "Yes\n";
                return 0;
            }
        }
    }
    cout << "No\n";
    return 0;
}

C - Reversible

只存字典序较小的

#include <bits/stdc++.h>

using namespace std;

#define int long long


int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
    int n;
    cin >> n;
    set<string> cnt;
    for (string s, t; n; n--) {
        cin >> s, t = s;
        reverse(s.begin(),s.end());
        s = min( s , t );
        cnt.insert(s);
    }
    cout << cnt.size() << "\n";
    return 0;
}

D - Peaceful Teams

直接暴搜，注意在搜索的过程中要及时盼到不合法的情况

#include <bits/stdc++.h>

using namespace std;

#define int long long

int n, T, m, res;
vector<vector<int>> g;
vector<set<int>> s;

void dfs(int x) {
    if (x > n) {
        if( s.size() == T ) res ++;
        return;
    }
    if( s.size() + n - x + 1 < T ) return;
    if (s.size() < T) {
        s.push_back(set<int>());
        s.back().insert(x);
        dfs(x + 1);
        s.pop_back();
    }
    for (auto &i: s) {
        int f = 1;
        for (auto j: g[x])
            if (i.count(j)) {
                f = 0;
                break;
            }
        if( f == 0 ) continue;
        i.insert(x);
        dfs( x + 1 );
        i.erase(x);
    }
    return;
}

int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
    cin >> n >> T >> m;
    g = vector<vector<int>>(n + 1);
    for (int x, y; m; m--) {
        cin >> x >> y;
        if (x < y) swap(x, y);
        g[x].push_back(y);
    }
    dfs(1);
    cout << res << "\n";
    return 0;
}

E - NAND repeatedly

简单的做一个 dp 就好了。

#include <bits/stdc++.h>

using namespace std;

#define int long long


int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
    int n , res = 0 ;
    string s;
    cin >> n >> s;
    vector<array<int,2>> f(n+1);
    for( int i = 1 , x ; i <= n ; i ++ ){
        x = s[i-1] - '0';
        if( x == 1 ) {
            f[i][1] = 1 + f[i-1][0];
            f[i][0] = f[i-1][1];
        }else{
            f[i][1] = f[i-1][0] + f[i-1][1];
            f[i][0] = 1;
        }
        res += f[i][1];
    }
    cout << res << '\n';
    return 0;
}

F - Make 10 Again

状压 dp，\(f[i][s]\)表示前\(i\)个数选部分数可以构成集合的概率。

对于枚举除了的\(s\)以及当前骰子掷出的数字\(j\)，有三种情况

s表示当前不要
1<<(j-1)表示之前的都不要
s<<j表示之前的加上当前的

把上述三种情况与到一起就是当前的数字全部可以构成的集合，注意要与\((1<<10)-1\)与一下防止溢出。

最后枚举所有包含10的集合把概率求和即为答案

#include <bits/stdc++.h>

using namespace std;

#define int long long

const int mod = 998244353;
const int N = 1 << 10;

int power(int x, int y) {
    int ans = 1;
    while (y) {
        if (y & 1) ans = ans * x % mod;
        x = x * x % mod, y >>= 1;
    }
    return ans;
}

int inv(int x) {
    return power(x, mod - 2);
}

int read() {
    int x = 0, f = 1, ch = getchar();
    while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
    if (ch == '-') f = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}

int32_t main() {
    int n = read();
    vector<int> f(N);
    f[0] = 1;
    for (int x, y, invs; n; n--) {
        x = read(), y = min(10ll, x), invs = inv(x);
        vector<int> g(N);
        for (int s = 0; s < N; s++) {
            for (int i = 1; i <= y; i++)
                (g[(s | s << i | 1 << (i - 1)) & (N - 1)] += f[s] * invs % mod) %= mod;
            (g[s] += f[s] * (x - y) % mod * invs % mod) %= mod;
        }
        swap(f, g);
    }
    int res = 0;
    for (int i = (1 << 9); i < N; i++)
        (res += f[i]) %= mod;
    cout << res << "\n";
    return 0;
}