Codeforces Round #697 (Div. 3)

Codeforces Round #697 (Div. 3)

A - Odd Divisor

int main() {
    IOS;
    for (cin >> _; _; --_) {
        ll n; cin >> n;
        while (n % 2 == 0) n /= 2;
        cout << (n > 1 && (n & 1) ? "YES\n" : "NO\n");
    }
    return 0;
}

B - New Year's Number

int main() {
    IOS;
    for (cin >> _; _; --_) {
        cin >> n;
        int c = n / 2020; n %= 2020;
        cout << (n <= c ? "YES\n" : "NO\n"); 
    }
    return 0;
}

C - Ball in Berland

计数, 去重

int main() {
    IOS;
    for (cin >> _; _; --_) {
        cin >> n >> m >> k; vector<PII> a(k); VI x(n + 1), y(m + 1);
        rep (i, 0, k - 1) cin >> a[i].fi;
        rep (i, 0, k - 1) cin >> a[i].se;
        sort(all(a)); a.erase(unique(all(a)), a.end());
        for (auto &i : a) ++x[i.fi], ++y[i.se];
        ll ans = 0;
        rep (i, 0, a.size() - 1) {
            --x[a[i].fi], --y[a[i].se];
            ans += a.size() - i - 1 - x[a[i].fi] - y[a[i].se];
        }
        cout << ans << '\n';
    }
    return 0;
}

D - Cleaning the Phone

按内存从大到小排序, 然后选择0~k-1 应用卸载, 并将0~k-1中的价值为2的选出来, 把剩下的价值为1的选出来

不断选择剩下最大内存的1去替换选中的2, 不断取价值最小值就行

int main() {
    IOS;
    for (cin >> _; _; --_) {
        cin >> n >> m; vector<PLL> a(n); multiset<PLL> st, y;
        rep(i, 0, n - 1) cin >> a[i].fi;
        rep(i, 0, n - 1) cin >> a[i].se;
        sort(all(a), greater<PLL>());
        ll k = 0, cur = 0, g = 0, mi;
        while (k < n && cur < m) {
            cur += a[k].fi, g += a[k].se;
            if (a[k].se == 2) st.insert(a[k]);
            ++k;
        }
        if (cur < m) { cout << "-1\n"; continue; } mi = g;
        for (int i = k; i < n && cur; ++i) if (a[i].se == 1) y.insert(a[i]);
        while (!st.empty()) {
            auto it = st.begin();
            cur -= it->fi, g -= it->se; st.erase(it);
            while (!y.empty() && cur < m) {
                auto mx = *y.rbegin();
                cur += mx.fi, g += mx.se;
                y.erase(y.find(mx));
            }
            if (cur < m) break;
            umin(mi, g);
        }
        cout << mi << '\n';
    }
    return 0;
}

E - Advertising Agency

求个组合数学

ll inv[N], fac[N], facinv[N];
 
int C(int m, int n) {
    return fac[n] * facinv[m] % mod * facinv[n - m] % mod;
}
 
void init() {
    inv[0] = inv[1] = fac[0] = fac[1] = facinv[0] = facinv[1] = 1;
    rep (i, 2, 2e5) {
        fac[i] = fac[i - 1] * i % mod;
        inv[i] = (mod - mod / i) * inv[mod % i] % mod;
        facinv[i] = facinv[i - 1] * inv[i] % mod;
    }
}
 
int main() {
    IOS; init();
    for (cin >> _; _; --_) {
        cin >> n >> k; VI a(n);
        for (auto &i : a) cin >> i;
        sort(all(a), greater<int>());
        int x = 0, y = 0;
        for (int i = k - 1; (~i) && a[k - 1] == a[i]; --i) ++x;
        for (int i = k; i < n && a[i] == a[k - 1]; ++i) ++y;
        cout << C(x, x + y) << '\n';
    }
    return 0;
}

F - Unusual Matrix

跟开关控制上下左右的题一样,

枚举第一行用不用异或1次, 之后, 第一行的数只能通过或列异或改变, 其他行的数只能通过行异或, 能选择的方式唯一, 判断是否可行即可

char a[N][N], b[N][N], d[N][N];
 
bool check() {
    rep (i, 2, n) {
        bool f = d[i][1] ^ b[i][1];
        rep (j, 2, n) if (d[i][j] ^ b[i][j] ^ f) return 0;
    }
    return 1;
}
 
int main() {
    IOS;
    for (cin >> _; _; --_) {
        cin >> n; bool f = 0;
        rep (i, 1, n) cin >> a[i] + 1;
        rep (i, 1, n) cin >> b[i] + 1;
        memcpy(d, a, sizeof a);
        rep (i, 1, n) if (d[1][i] != b[1][i]) rep (j, 1, n) d[j][i] = d[j][i] == '1' ? '0' : '1';
        f = check();
        memcpy(d, a, sizeof a); rep (i, 1, n) d[1][i] = d[1][i] == '1' ? '0' : '1';
        rep (i, 1, n) if (d[1][i] != b[1][i]) rep (j, 1, n) d[j][i] = d[j][i] == '1' ? '0' : '1';
        f = f | check();
        cout << (f ? "YES\n" : "NO\n");
    }
    return 0;
}

G - Strange Beauty

枚举倍数就好了, nlogn, t <= 10, 总复杂度\(O(tnlogn)\)

int main() {
    IOS;
    for (cin >> _; _; --_) {
        cin >> n; VI a(2e5 + 1, 0), c(2e5 + 1, 0);
        for (int i = 0; i < n; ++i) cin >> m, ++c[m];
        for (int i = 1; i <= 2e5; ++i) if (c[i]) {
            a[i] += c[i];
            for (int j = i * 2; j <= 2e5; j += i) if (c[j]) umax(a[j], a[i]);
        }
        cout << n - (*max_element(all(a))) << '\n';
    }
    return 0;
}