https://ac.nowcoder.com/acm/contest/91355
A

#include<bits/stdc++.h>
using namespace std;
signed main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int T;
    cin >> T;
    while (T--) {
        vector<int> a(101);
        int n, k;
        cin >> n >> k;
        for (int i = 1; i <= n; i++) {
            int x;
            cin >> x;
            a[x]++;
        }
        for (int i = 1; i <= k; i++) {
            if (a[i] == 0) {
                cout << "NO" << endl;
                goto end;
            }
        }
        cout << "YES" << endl;
        end:;
    }
    return 0;
}

B

#include<bits/stdc++.h>
using namespace std;
signed main() {
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T;
    cin >> T;
    cout << fixed << setprecision(15);
    while (T--) {
        double a;
        cin >> a;
        cout << (a + sqrtl(a * a + 4)) / 2 << '\n';
    }
    return 0;
}

C
将数组排序 如若和大于sum则从后往前修改，否则从前往后修改模拟即可

#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
signed main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int T;
    cin >> T;
    while (T--) {
        int n, sum;
        cin >> n >> sum;
        int sum0 = 0;
        vector<int> a(n + 1);
        for (int i = 1; i <= n; i++)cin >> a[i], sum0 += a[i];
        int ans = 0;
        if (sum0 == sum)cout << 0 << endl;
        else if (sum0 > sum) {//把大数边小数
            sort(a.begin() + 1, a.end());
            for (int i = n; i >=1; i--) {
                if (sum0 - a[i] - 10000 > sum) {
                    ans++;
                    sum0 -= a[i] + 10000;
                } else {
                    ans++;
                    break;
                }
            }
            cout << ans << endl;
        } else {
            std::sort(a.begin() + 1, a.end());
            for (int i = 1; i <=n; i++) {
                if (sum0 - a[i] + 10000 < sum) {
                    ans++;
                    sum0 += 10000 - a[i];
                } else {
                    ans++;
                    break;
                }
            }
            cout << ans << endl;
        }
    }
    return 0;
}

D
分层图 思考方式类dp 考虑使用，dp[i][j]表示走到i,j天没有休息的最短时间。
在路径转移中有两种操作
休息与不休息两种方式
不休息时需满足j+1<=k转移到j+1层
休息时满足转移至第0层

#include<bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'
const int N = 200000;
int dist[N + 1][11];
int st[N + 1][11];

signed main() 
{
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int T;
    cin >> T;
    while (T--) {
        int n, m, k;
        cin >> n >> m >> k;
        vector<int> a(n + 1);
        for (int i = 1; i <= n; i++)cin >> a[i];
        vector<vector<int>> g(n + 1);
        for (int i = 1; i <= m; i++) {
            int u, v;
            cin >> u >> v;
            g[u].push_back(v);
            g[v].push_back(u);
        }
        for (int i = 1; i <= n; i++) 
        {
            for (int j = 0; j <= k; j++) 
            {
                dist[i][j] = 1e18;
                st[i][j] = 0;
            }
        }
        priority_queue<array<int, 3>, vector<array<int, 3>>, greater<>> q;
        if (k != 0) 
        {
            dist[1][1] = 1;
            q.push({1, 1, 1});
        }
        dist[1][0] = a[1];
        q.push({a[1], 1, 0});
        while (q.size()) 
        {
            auto [w, i, j] = q.top();
            q.pop();
            if (st[i][j])continue;
            st[i][j] = 1;
            for (auto ne: g[i]) 
            {
                //休息
                if (dist[ne][0] > a[ne] + w) 
                {
                    dist[ne][0] = a[ne] + w;
                    q.push({dist[ne][0], ne, 0});
                }
                //不休息
                if (j + 1 <= k && dist[ne][j + 1] > w + 1)
                {
                    dist[ne][j + 1] = w + 1;
                    q.push({dist[ne][j + 1], ne, j + 1});
                }
            }
        }
        int ans = 1e18;
        for (int i = 0; i <= k; i++)ans = min(ans, dist[n][i]);
        cout << ans << endl;
    }
    return 0;
}