Codeforces Round 900 (Div. 3) 题解 | JorbanS

A-How Much Does Daytona Cost?

string solve() {
    cin >> n >> m;
    bool flag = false;
    while (n --) {
        int x; cin >> x;
        if (x == m) flag = true;
    }
    return flag ? yes : no;
}
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B-Aleksa and Stack

void solve() {
    cin >> n;
    int a = 2, b = 3;
    while (n --) {
        cout << a << ' ';
        int x = b + 1;
        while (3 * x % (a + b) == 0) x ++;
        a = b, b = x;
    }
    cout << endl;
}
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C-Vasilije in Cacak

题解 只要 $x$ 在最小值和最大值区间内，$x$ 就一定能被满足，部分可以用 $RMQ$ 来理解

string solve() {
    cin >> n >> k >> x;
    ll minn = (1 + k) * k / 2, maxx = (n * 2 - k + 1) * k / 2;
    return x < minn || x > maxx ? no : yes;
}
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D-Reverse Madness

题解 对于每个区间分别处理，二分分类区间，每个区间用差分处理

string solve() {
    cin >> n >> k >> s;
    for (int i = 1; i <= n; i ++) a[i] = pre[i] = 0;
    for (int i = 1; i <= k; i ++) cin >> l[i];
    for (int i = 1; i <= k; i ++) cin >> r[i], lr[i] = l[i] + r[i];
    cin >> q;
    vector<vector<int>> v(n + 1);
    for (int i = 1; i <= q; i ++) {
        int x; cin >> x;
        int p = lower_bound(r + 1, r + k + 1, x) - r;
        v[p].push_back(min(x, lr[p] - x));
    }
    for (int i = 1; i <= k; i ++) {
        a[l[i]] = pre[l[i] - 1] = 0;
        for (auto L : v[i]) a[L] ++, a[lr[i] - L + 1] --;
        for (int j = l[i]; j <= lr[i] / 2; j ++) pre[j] = pre[j - 1] + a[j];
    }
    for (int i = 1; i <= k; i ++)
        for (int j = l[i]; j <= lr[i] / 2; j ++)
            if (pre[j] % 2) swap(s[j - 1], s[lr[i] - j - 1]);
    return s;
}
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E-Iva & Pav

题解 $RMQ$，但是因为 $D$ 题浪费太多时间了，没时间写了（

赛后补了后面的查询部分，一下 $AC$ 了

void solve() {
    cin >> n;
    vector<vector<int>> to(n + 2, vector<int>(19));
    for (int i = 1; i <= n; i ++) cin >> f[i][0], to[i][0] = i + 1;
    for (int i = 1; i <= 17; i ++) {
        for (int j = 1; j <= n; j ++) {
            if (!to[to[j][i - 1]][i - 1]) break;
            to[j][i] = to[to[j][i - 1]][i - 1];
            f[j][i] = f[j][i - 1] & f[to[j][i - 1]][i - 1];
        }
    }
    cin >> m;
    while (m --) {
        int l, k; cin >> l >> k;
        int r = l, now = f[l][0];
        if (now < k) {
            cout << -1 << ' ';
            continue;
        }
        for (int i = 17; i >= 0; i --) {
            if (to[r][i] && (now & f[r][i]) >= k) {
                now &= f[r][i];
                r = to[r][i];
            }
        }
        cout << r - 1 << ' ';
    }
    cout << endl;
}
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F - Vasilije Loves Number Theory

题解 对于 $n=p_1^{x_1}\cdot p_2^{x_2}\dots$ 而言，$d(n)=(x_1+1)\cdot (x_2+1)\dots$

对于两数 $a,b$ 有$gcd(a,b)=1$ 时，$d(n)$ 是个积性函数

因为要判断 $d(a\cdot n)=n$ 的 $a$ 是否存在，而且有 $gcd(a,n)=1$，说明 $n$ 乘上 $a$ 计算 $d(a\cdot n)$ 值时，$a$ 不能修改 $n$ 原有因子的数量，因此 $d(a\cdot n)=d(a)\cdot d(n)$，也就是说只要 $d(n)|n$ 就一定能找到 $a$

由于题目数据范围 $d(n)\le 10^9$，故 $n$ 会爆 $longlong$

所以不能直接判断 $n\%d(n)$ 是否为 $0$ 来判断，因此对 $\prod n$ 逐步除以 $gcd(n,d(\prod n))$ 即可，若结果为 $1$ 则符合

vector<int> primes;

void getPrimes() {
    vector<bool> vis(1e6 + 1);
    for (int i = 2; i <= 1e6; i ++) {
        if (!vis[i]) primes.push_back(i);
        for (int x : primes) {
            if (x * i > 1e6) break;
            vis[x * i] = true;
            if (x % i == 0) break;
        }
    }
}

int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }

void solve() {
    int n, m; cin >> n >> m;
    map<int, int> mp, mpN;
    auto getDivisors = [&](int x) -> void {
        for (auto i : primes) {
            if (i > x) break;
            while (x % i == 0) mp[i] ++, x /= i;
        }
        if (x != 1) mp[x] ++;
    };
    getDivisors(n);
    mpN = mp;
    vector<int> a;
    a.push_back(n);
    while (m --) {
        int op; cin >> op;
        if (op == 1) {
            int x; cin >> x;
            getDivisors(x);
            int res = 1;
            for (auto [u, v] : mp) res *= v + 1;
            a.push_back(x);
            for (auto it : a) res /= gcd(res, it);
            cout << (res == 1 ? yes : no) << endl;
        } else {
            mp = mpN;
            a.clear();
            a.push_back(n);
        }
    }
    cout << endl;
}

int main() {
    FastIO
    getPrimes();
    Cases
    solve();
    return 0;
}
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