A - tcdr

#include <bits/stdc++.h>

using namespace std;

int32_t main() {
    string s;
    cin >> s;
    for( auto i : s){
        if( i != 'a' and i != 'e' and i != 'i' and i != 'o' and i != 'u' )
            cout << i;
    }
    return 0;
}

B - The Middle Day

#include <bits/stdc++.h>

using namespace std;

#define int long long

int32_t main() {
    int n, sum = 0;
    cin >> n;
    vector<int> a(n + 1);
    for (int i = 1; i <= n; i++) cin >> a[i], sum += a[i];
    sum = sum / 2 + 1;
    for (int i = 1; i <= n; i++) {
        if (sum > a[i])sum -= a[i];
        else {
            cout << i << " " << sum << "\n";
            return 0;
        }
    }
    return 0;
}

C - Flavors

#include <bits/stdc++.h>

using namespace std;

#define int long long

int32_t main() {
    int n ;
    cin >> n;
    map<int,vector<int>> a;
    for( int f , s ; n ; n -- )
        cin >> f >> s, a[f].push_back(s);
    vector<int> b;
    int res = 0;
    for( auto &[k,v] : a ){
        sort(v.begin(), v.end() , greater<int>());
        b.push_back(v[0]);
        if( v.size() >= 2 ) res = max( res , v[0] + v[1] / 2 );
    }
    sort( b.begin(), b.end() , greater<int>() );

    if( b.size() >= 2 ) res = max( res , b[0] + b[1] );
    cout << res << "\n";

    return 0;
}

D - Magical Cookies

赛时被这道题卡住了很久。

一开始写了一个暴力的代码但是把复杂度估计错了。

#include <bits/stdc++.h>

using namespace std;

#define int long long

int32_t main() {
    int n, m;
    cin >> n >> m;
    vector<string> f(n), g;
    for (auto &i: f) cin >> i;
    for (bool flag; n > 0 and m > 0;) {
        flag = true;
        set<int> r , c;
        for (int i = 0, t; i < n and m >= 2 ; i++) {
            t = 1;
            for (int j = 1; t and j < m; j++)
                if (f[i][j] != f[i][0]) t = 0;
            if (t == 0) continue;
            r.insert(i);
        }
        for (int j = 0, t; j < m and n >= 2 ; j++) {
            t = 1;
            for (int i = 1; t and i < n; i++)
                if (f[i][j] != f[0][j]) t = 0;
            if (t == 0) continue;
            c.insert(j);
        }
        if( r.empty() and c.empty() ) break;
        g = vector<string>();
        for( int i = 0 ; i < n ; i ++ ){
            if( r.count(i) ) continue;
            g.push_back("");
            for( int j = 0 ; j < m ; j ++ ){
                if( c.count(j) ) continue;
                g.back() += f[i][j];
            }
        }
        n = g.size();
        if(n) m = g.front().size();
        f = g;
    }
    cout << n * m << "\n";
    return 0;
}

认真读题后，发现删除行列只能是整行整列的形式，所以剩下的部分始终是一个矩形。并且我们实际上并不关注行列的字母排列，只需要知道行列中字母的数量。所以我们维护每一行列中字母的数量即可。

#include <bits/stdc++.h>

using namespace std;

#define int long long

int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int n, m;
    cin >> n >> m;
    vector<string> f(n);
    for (auto &i: f) cin >> i;
    vector<array<int, 26>> R(n), C(m);
    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++)
            R[i][f[i][j] - 'a']++, C[j][f[i][j] - 'a']++;
    for (int x, y; n > 0 and m > 0;) {
        vector<pair<int, int>> r, c;
        for (int i = 0; i < R.size() and m >= 2 ; i++)
            for (int j = 0; j < 26; j++)
                if (R[i][j] == m) r.emplace_back(i, j);

        for (int i = 0; i < C.size() and n >= 2 ; i++)
            for (int j = 0; j < 26; j++)
                if (C[i][j] == n) c.emplace_back(i, j);
        if( r.empty() and c.empty() ) break;
        for( auto [ rr , ch] : r ){
            R[rr][ch] = 0;
            for( int i = 0 ; i < C.size() ; i ++ )
                C[i][ch] = max( 0ll , C[i][ch] - 1);
        }
        for( auto [cc,ch] : c ){
            C[cc][ch] = 0;
            for( int i = 0 ; i < R.size() ; i ++ )
                R[i][ch] = max(0ll , R[i][ch] - 1);
        }
        n -= r.size() , m -= c.size();

    }
    cout << n * m << "\n";
    return 0;
}

E - Prerequisites

反向建图，然后从\(1\)开始dfs 求出生成树，生成树上点即为必须要读的前驱。然后统计前驱中入读为零的点即可。

#include <bits/stdc++.h>

using namespace std;

#define int long long

int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int n;
    cin >> n;
    vector<vector<int>> e(n + 1), g(n + 1);
    vector<int> vis(n + 1), inner(n + 1);
    for (int i = 1, x; i <= n; i++) {
        cin >> x;
        e[i].resize(x);
        for (auto &y: e[i]) cin >> y;
    }

    auto dfs = [&g, &vis, &inner, e](auto &&self, int x) {
        if (vis[x]) return;
        vis[x] = 1;
        for (auto y: e[x]) {
            g[y].push_back(x), inner[x]++;
            self(self, y);
        }
        return ;
    };

    dfs(dfs, 1);
    
    vector<int> res;
    queue<int> q;
    for (int i = 1; i <= n; i++)
        if (vis[i] and inner[i] == 0) q.push(i);
    while (!q.empty()) {
        auto x = q.front();
        q.pop();
        res.push_back(x);
        for (auto y: g[x]) {
            inner[y]--;
            if (inner[y] == 0) q.push(y);
        }
    }

    res.pop_back();
    for (auto i: res)
        cout << i << " ";
    return 0;
}

F - Shortcuts

可以知道的是能够跳过的点的数量其实不多。估算后大约最多只能跳过30个点，因为跳的点足够多之后反而不如直接走更优。

那么这样的就可以dp，状态\(f[i][j]\)表示走到点\(i\)跳过\(j\)个点的最优解。然后可以直接枚举转移到当前状态跳过了多少个点\(k\)，则转移为\(f[i][j]=\min(f[i][j],f[i-k-1][j-k]) + dis(i,i-k-1)+dirt(j , j - k )\)，其中\(dis\)表示两点的距离，\(dirt\)是先减去跳过\(j-k\)个点的惩罚再加上\(j\)个点的惩罚。

#include <bits/stdc++.h>

using namespace std;

#define int long long
#define double long double
typedef pair<double, double> pdd;

const double inf = 1e10;

double dis(pdd a, pdd b) {
    return sqrt((a.first - b.first) * (a.first - b.first) + (a.second - b.second) * (a.second - b.second));
}

double dirt(int x, int y) {
    double ans;
    if (x > 0) ans = (1ll << (x - 1));
    else ans = 0;
    if (y > 0) ans -= (1ll << (y - 1));
    return ans;
}

int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int n;
    cin >> n;
    vector<pdd> d(n + 1);
    for (int i = 1; i <= n; i++)
        cin >> d[i].first >> d[i].second;
    vector f(n + 1, vector(35, (double) (inf)));
    f[1][0] = 0;
    for (int x = 2; x <= n; x++)
        for (int j = 0; j < 35; j++)
            for (int k = 0, y = x - 1; y >= 1 and k <= j; k++, y--)
                f[x][j] = min(f[x][j], f[y][j - k] + dis(d[x], d[y]) + dirt( j , j - k ));

    cout << fixed << setprecision(10) << *min_element(f[n].begin(), f[n].end()) << "\n";
    return 0;
}

G - Ai + Bj + Ck = X (1 <= i, j, k <= N)

枚举\(i\)的值，然后方程就变成了\(Bj+Ck=X-Ai\)。用扩欧解丢番图方程即可。

注意本体精度问题。

#include <bits/stdc++.h>

using namespace std;

#define int __int128

typedef long long ll;

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;
}

int exgcd(int a, int b, int &x, int &y) {
    if (b == 0) {
        x = 1, y = 0;
        return a;
    }
    int d = exgcd(b, a % b, x, y);
    int z = x;
    x = y, y = z - y * (a / b);
    return d;
}

template<typename T, typename U>
T ceil(T x, U y) {
    return (x > 0 ? (x + y - 1) / y : x / y);
}

template<typename T, typename U>
T floor(T x, U y) {
    return (x > 0 ? x / y : (x - y + 1) / y);
}

int32_t main() {
    int n = read(), a = read(), b = read(), c = read(), x = read(), res = 0;
    for (int i = 1, j, k, m, d, bp, cp, l, r; i <= n; i++) {
        m = x - a * i;
        d = exgcd(b, c, j, k);
        if (m % d) continue;
        j *= m / d, k *= m / d;
        bp = b / d, cp = c / d;
        l = max(ceil(1 - j, cp), ceil(k - n, bp));
        r = min(floor(n - j, cp), floor(k - 1, bp));
        if (r >= l) res += r - l + 1;
    }
    cout << (ll) res << "\n";
    return 0;
}