Codeforces Round 977 (Div. 2, based on COMPFEST 16 - Final Round)

A

不断取最小的两个去凑即可。

#include <bits/stdc++.h>
using namespace std;
void solve()
{
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; i++)
    {
        cin >> a[i];
    }
    while (a.size() > 1)
    {
        sort(a.begin(), a.end());
        a[0] = (a[0] + a[1]) >> 1;
        a.erase(a.begin() + 1);
    }
    cout << a[0] << '\n';
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int t;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

B

一种不断凑的方法。

#include <bits/stdc++.h>
using namespace std;
void solve()
{
    int n, m;
    cin >> n >> m;
    vector<int> a(n);
    for (int i = 0; i < n; i++)
    {
        int x;
        cin >> x;
        if (x < n)
        {
            a[x]++;
        }
    }
    for (int i = 0; i < n; i++)
    {
        if (a[i] == 0)
        {
            cout << i << '\n';
            return;
        }
        a[i]--;
        if (i + m < n)
        {
            a[i + m] += a[i];
        }
    }
    cout << n << '\n';
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int t;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

C

按照顺序依次就是正确的。

#include <bits/stdc++.h>
using namespace std;
void solve()
{
    int n, m, q;
    cin >> n >> m >> q;
    vector<int> a(n);
    vector<int> b(m);
    map<int, int> mp;
    for (int i = 0; i < n; i++)
    {
        cin >> a[i];
    }
    int cnt = 0;
    for (int i = 0; i < m; i++)
    {
        int x;
        cin >> x;
        if (!mp[x])
        {
            b[cnt++] = x;
        }
        mp[x] = 1;
    }
    for (int i = 0; i < cnt; i++)
    {
        if (a[i] != b[i])
        {
            cout << "TIDAK" << '\n';
            return;
        }
    }
    cout << "YA" << '\n';
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int t;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}