A - Filter

#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;
    for( int x ; n ; n -- ){
        cin >> x;
        if( x & 1 ) continue;
        cout << x << " ";
    }
    return 0;
}

B - ASCII Art

#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;
    for( int i = 1 , x ; i <= n ; i ++ ){
        for( int j = 1 ; j <= m ; j ++ ){
            cin >> x;
            if( x == 0 ) cout << ".";
            else cout << (char)('A'+x-1);
        }
        cout << "\n";
    }
    return 0;
}

C - Merge Sequences

模拟一下归并排序的过程

#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> a(n), b(m), c, d;
    for (auto &i: a)
        cin >> i;
    for (auto &i: b)
        cin >> i;
    for (int t = 1, i = 0, j = 0; t <= n + m; t++) {
        if (i == n) d.push_back(t), j++;
        else if (j == m) c.push_back(t), i++;
        else if (a[i] <= b[j]) c.push_back(t), i++;
        else d.push_back(t), j++;
    }
    for (auto i: c)
        cout << i << " ";
    cout << "\n";
    for (auto i: d)
        cout << i << " ";
    cout << "\n";

    return 0;
}

D - Bank

用两个set模拟一下这个过程就好了。

#include <bits/stdc++.h>

using namespace std;

#define int long long

int cnt[26];

int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
    int n, q;
    cin >> n >> q;
    set<int> a, b;
    for (int i = 1; i <= n; i++)
        a.insert(i);
    for (int op, x; q; q--) {
        cin >> op;
        if (op == 1)
            b.insert(*a.begin()), a.erase(*a.begin());
        else if( op == 2 ){
            cin >> x;
            b.erase(x);
        }else
            cout << *b.begin() << "\n";
    }
    return 0;
}

E - 2xN Grid

这个双指针就可以做

#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, na, nb, res = 0;
    cin >> n >> na >> nb;
    vector<pair<int, int>> a(na);
    for (auto &[x, y]: a)
        cin >> x >> y;
    for( int i = 1 , pa = 0 , v , l  ; i <= nb ; i ++ ){
        cin >> v >> l;
        while( l ){
            int len = min( l , a[pa].second );
            if( v == a[pa].first ) res += len;
            a[pa].second -= len , l -= len;
            if( a[pa].second == 0 ) pa ++;
        }
    }
    cout << res << "\n";
    return 0;
}

F - Sugar Water 2

我们的做法就是二分答案。二分第\(k\)大的浓度是\(x\)，然后统计有多少种情况混合浓度大于\(x\)

浓度大于\(x\)的情况是$\frac{A_i+C_j}{A_i+B_i+C_j+D_j} > x \(，化简可以得到\)(1-x)A_i-B_ix>(x-1)C_j+D_jx$

发现不等式两边互补影响，我们算出两种糖水的参数\(p_i=(1-x)A_i-B_ix,q_i=(x-1)C_j+D_jx\)

把\(p,q\)排序后用双指针算出有多少对\((i,j)\)满足\(p_i>q_j\)即可，复杂度\(O(\log10^{12}N\log N)\)

#include <bits/stdc++.h>

using namespace std;

#define int long long
#define double long double

int n, m, k;
vector<double> a, b, c, d;

constexpr double eps = 1e-12;

bool check(double x) {
    int cnt = 0;
    vector<double> p, q;
    for (int i = 0; i < n; i++)
        p.push_back((1.0 - x) * a[i] - b[i] * x);
    for (int i = 0; i < m; i++)
        q.push_back((x - 1.0) * c[i] + d[i] * x);
    sort(p.begin(), p.end()), sort(q.begin(), q.end());
    for (int i = 0, j = 0; i < n; i++) {
        while (j < m && q[j] < p[i]) j++;
        cnt += j;
    }
    return cnt >= k;
}

int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
    cin >> n >> m >> k;
    a = vector<double>(n), b = vector<double>(n);
    c = vector<double>(m), d = vector<double>(m);
    for (int i = 0; i < n; i++)
        cin >> a[i] >> b[i];
    for (int i = 0; i < m; i++)
        cin >> c[i] >> d[i];

    double l = 0, r = 1, mid, res;
    while (r - l > eps) {
        mid = (l + r) / 2.0;
        if (check(mid)) res = mid, l = mid + eps;
        else r = mid - eps;
    }
    cout << fixed << setprecision(9) << res * 100.0;
    return 0;
}