Educational Codeforces Round 108 (Rated for Div. 2)

Educational Codeforces Round 108 (Rated for Div. 2)

A - Red and Blue Beans

int main() {
    IOS;
    for (cin >> _; _; --_) {
        cin >> n >> m >> k;
        if (n > m) swap(n, m);
        m -= n;
        cout << (!m || (m - 1) / n + 1 <= k ? "YES\n" : "NO\n");
    }
    return 0;
}

B - The Cake Is a Lie

int main() {
    IOS;
    for (cin >> _; _; --_) {
        cin >> n >> m >> k;
        if (n < m) swap(n, m);
        k -= n - 1 + (m - 1ll) * n;
        cout << (!k ? "YES\n" : "NO\n");
    }
    return 0;
}

C - Berland Regional

排序之后对每个学校算对每种人数组队的贡献即可

int u[N];
ll s[N];
 
int main() {
    IOS;
    for (cin >> _; _; --_) {
        cin >> n; map<int, vector<ll>> st;
        rep (i, 1, n) cin >> u[i], s[i] = 0;
        rep (i, 1, n) cin >> k, st[u[i]].pb(k);
        for (auto &i : st) {
            sort(all(i.se), greater<int>());
            rep (j, 1, i.se.size() - 1) i.se[j] += i.se[j - 1];
            rep (j, 1, i.se.size()) s[j] += i.se[i.se.size() / j * j - 1];
        }
        rep (i, 1, n) cout << s[i] << char(" \n"[i == n]);
    }
    return 0;
}

D - Maximum Sum of Products

区间dp模型

int a[N], b[N];
ll sum, mx, s[N][N];
 
int main() {
    IOS; cin >> n;
    rep (i, 1, n) cin >> a[i];
    rep (i, 1, n) cin >> b[i], sum += (ll)a[i] * b[i];
    rep (i, 2, n) rep (j, 1, n + 1 - i)
        umax(mx, s[j][i + j - 1] = s[j + 1][i + j - 2] + (ll)(a[i + j - 1] - a[j]) * (b[j] - b[i + j - 1]));
    cout << (sum + mx);
    return 0;
}