Educational Codeforces Round 167 (Rated for Div. 2)

Educational Codeforces Round 167 (Rated for Div. 2)

A. Catch the Coin

解题思路：

最终\(x\)一定会相等，我们考虑直接到下面接住他。

代码：

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using i128 = __int128_t;
typedef pair<int, int> pii;
#define fi first
#define se second
const int N = 1e5 + 10;

void solve()
{
    int a, b;
    cin >> a >> b;
    if (a == 0 && b == -1)
    {
        cout << "YES\n";
        return;
    }
    int c = -abs(a);
    b -= abs((abs(a) - 1));
    if (b >= c)
    {
        cout << "YES\n";
    }
    else
    {
        cout << "NO\n";
    }
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

B. Substring and Subsequence

解题思路:

双循环暴力查找最优添加法。

代码:

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using i128 = __int128_t;
typedef pair<int, int> pii;
#define fi first
#define se second
const int N = 1e5 + 10;

void solve()
{
    string a, b;
    cin >> a >> b;
    int ans = 0;
    for (int i = 0; i < b.size(); i++)
    {
        int idx = i;
        int cnt = 0;
        for (int j = 0; j < a.size(); j++)
        {
            if (a[j] == b[idx])
            {
                idx++;
                cnt++;
            }
        }
        ans = max(ans, cnt);
    }
    int res = a.size() + (b.size() - ans);
    cout << res << "\n";
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

C. Two Movies

解题思路：

分情况讨论。

如果\(a \neq b:\)选大的那个。

如果\(a = b:\)优先让小的变大，然后让大的变小，最后均分。

代码：

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using i128 = __int128_t;
typedef pair<int, int> pii;
#define fi first
#define se second
const int N = 1e5 + 10;

void solve()
{
    int n;
    cin >> n;
    vector<int> a(n + 1), b(n + 1);
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
    }
    for (int i = 1; i <= n; i++)
    {
        cin >> b[i];
    }
    vector<int> cnt(2), cur(2);
    for (int i = 1; i <= n; i++)
    {
        if (a[i] == b[i])
        {
            if (a[i] == 1)
            {
                cnt[0]++;
            }
            else if (a[i] == -1)
            {
                cnt[1]++;
            }
        }
        else
        {
            if (a[i] == 1)
            {
                cur[0]++;
            }
            else if (b[i] == 1)
            {
                cur[1]++;
            }
        }
    }
    // cout << cur[0] << ' ' << cur[1] << endl;
    // cout << cnt[0] << ' ' << cnt[1] << endl;
    if (cur[0] > cur[1])
    {
        swap(cur[0], cur[1]);
    }
    while (cur[0] < cur[1] && cnt[0] > 0)
    {
        cnt[0]--;
        cur[0]++;
    }
    while (cur[0] < cur[1] && cnt[1] > 0)
    {
        cur[1]--;
        cnt[1]--;
    }
    int res = cnt[0] - cnt[1];
    // cout << cur[0] << ' ' << cur[1] << ' ';
    // cout << res << endl;
    if (res > 0)
    {
        res = res / 2;
        if (cur[0] == cur[1])
        {
            cout << cur[0] + res << '\n';
        }
        else
        {
            cout << cur[0] << '\n';
        }
    }
    else
    {
        res = (res - 1) / 2;
        if (cur[0] == cur[1])
        {
            cout << cur[0] + res << '\n';
        }
        else
        {
            cout << cur[0] << '\n';
        }
    }
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

D. Smithing Skill

解题思路：

双关键词升序排序：第一关键字\(a_i - b_i\),第二关键字\(a_i\)。

\(f_i:花费i个万能金属能得到的最大经验值。\)

考虑数据的单调性，双指针预处理\(f_i\)。

对于\(c_i > 10^6\)要处理到\(10^6\)以下。

代码：

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using i128 = __int128_t;
typedef pair<int, int> pii;
#define fi first
#define se second
const int N = 1e6;

void solve()
{
    ll n, m;
    cin >> n >> m;
    vector<ll> a(n + 1), b(n + 1), c(m + 1);
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
    }
    for (int i = 1; i <= n; i++)
    {
        cin >> b[i];
    }
    for (int i = 1; i <= m; i++)
    {
        cin >> c[i];
    }
    vector<pii> v(1);
    for (int i = 1; i <= n; i++)
    {
        ll t = a[i] - b[i];
        v.push_back({t, a[i]});
        // cout << t << ' ' << a[i] << endl;
    }
    sort(v.begin() + 1, v.end());
    vector<pii> d(1);
    for (int i = 1; i <= n; i++)
    {
        if (d.size() > 1)
        {
            if (v[i].se < d.back().se)
            {
                d.push_back(v[i]);
            }
        }
        else
        {
            d.push_back(v[i]);
        }
    }
    // for (int i = 1; i < d.size(); i++)
    // {
    //     cout << d[i].fi << ' ' << d[i].se << endl;
    // }
    vector<int> f(N + 10);
    for (int i = 1, j = d.size(); i <= N; i++)
    {
        while ((j > 1) && d[j - 1].se <= i)
        {
            j--;
        }
        if (j != (int)d.size())
        {
            auto [x, y] = d[j];
            f[i] = f[i - x] + 1;
        }
    }
    ll ans = 0;
    for (int i = 1; i <= m; i++)
    {
        if (c[i] > N)
        {
            int k = (c[i] - d[1].se + (d[1].fi - 1)) / d[1].fi;
            c[i] -= k * d[1].fi;
            ans += k;
        }
        ans += f[c[i]];
    }
    cout << ans * 2 << '\n';
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    // cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}