第 380 场 LeetCode 周赛题解

A 最大频率元素计数

模拟：先统计元素的频率，然后求由最大频率的元素的总频率

class Solution {
public:
    int maxFrequencyElements(vector<int> &nums) {
        unordered_map<int, int> cnt;
        for (auto x: nums)
            cnt[x]++;
        int mx = 0, s = 0;
        for (auto [_, f]: cnt)
            if (f > mx) {
                mx = f;
                s = f;
            } else if (f == mx)
                s += f;
        return s;
    }
};

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B 找出数组中的美丽下标 I

同t4…

class Solution {
public:
    vector<int> beautifulIndices(string s, string a, string b, int k) {
        srand(time(0));//随机种子
        int e = 2333 + rand() % 100, mod = 1e9 + rand() % 100;
        shash hs(s, e, mod), ha(a, e, mod), hb(b, e, mod);
        vector<int> vi, vj;
        int ns = s.size(), na = a.size(), nb = b.size();
        for (int i = 0; i <= ns - na; i++)
            if (hs(i, i + na - 1) == ha(0, na - 1))
                vi.push_back(i);
        for (int j = 0; j <= ns - nb; j++)
            if (hs(j, j + nb - 1) == hb(0, nb - 1))
                vj.push_back(j);
        vector<int> res;
        for (auto i: vi) {
            auto l = lower_bound(vj.begin(), vj.end(), i - k);
            auto r = lower_bound(vj.begin(), vj.end(), i + k + 1);
            if (l != r)
                res.push_back(i);
        }
        return res;
    }

    class shash {//字符串哈希模板
    public:
        using ll = long long;
        vector<ll> pres;
        vector<ll> epow;
        ll e, p;

        shash(string &s, ll e, ll p) {
            int n = s.size();
            this->e = e;
            this->p = p;
            pres = vector<ll>(n + 1);
            epow = vector<ll>(n + 1);
            epow[0] = 1;
            for (int i = 0; i < n; i++) {
                pres[i + 1] = (pres[i] * e + s[i]) % p;
                epow[i + 1] = (epow[i] * e) % p;
            }
        }

        ll operator()(int l, int r) {
            ll res = (pres[r + 1] - pres[l] * epow[r - l + 1] % p) % p;
            return (res + p) % p;
        }
    };
};

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C 价值和小于等于 K 的最大数字

二分 + 数学：通过二分枚举答案，这需要求不超过 $mid$ 的非负数（ $0$ 的价值和为 $0$ 不影响答案）的价值和，枚举二进制中的满足 $i\%x==0$ 的第 $i$ 位， $0～mid$ 的第 $i$ 位形成长为 $2^i$ 的循环数列的前 $mid+1$ 项：$\underset{2^{i?1}个0}{\underbrace{0,?,0}},\underset{2^{i?1}个1}{\underbrace{1,?,1}}?$，所以可以直接求 $0～mid$ 中第 $i$ 位为 $1$ 的数的数目

class Solution {
public:
    using ll = long long;

    long long findMaximumNumber(long long k, int x) {
        ll l = 1, r = 1e15;
        while (l < r) {
            ll mid = (l + r + 1) / 2;
            ll s = 0;
            for (int i = 0; i < 50; i++)
                if ((i + 1) % x == 0) {
                    ll e = 1LL << (i + 1), se = 1LL << i;
                    ll c = (mid + 1) / e, mod = (mid + 1) % e;
                    s += c * se + (mod > se ? mod - se : 0);
                }
            if (s <= k)
                l = mid;
            else
                r = mid - 1;
        }
        return l;
    }
};

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D 找出数组中的美丽下标 II

字符串哈希 + 二分：先用字符串哈希求出满足 $0\le i\le s.length?a.length$ 和 $s[i,(i+a.length?1)]=a$ 的 $i$ 构成的有序数组 $vi$，和满足 $0\le j\le s.length?b.length$ 和 $s[j,(j+b.length?1)]=b$ 的 $j$ 构成的有序数组 $vj$ ，然后枚举 $vi$ 中的 $i$ ，在 $vj$ 中二分查找 $j\in [i?k,i+k?1)$ ，如果存在这样的 $j$ ，则将 $i$ 加入答案数组

class Solution {
public:
    vector<int> beautifulIndices(string s, string a, string b, int k) {
        srand(time(0));//随机种子
        int e = 2333 + rand() % 100, mod = 1e9 + rand() % 100;
        shash hs(s, e, mod), ha(a, e, mod), hb(b, e, mod);
        vector<int> vi, vj;
        int ns = s.size(), na = a.size(), nb = b.size();
        for (int i = 0; i <= ns - na; i++)
            if (hs(i, i + na - 1) == ha(0, na - 1))
                vi.push_back(i);
        for (int j = 0; j <= ns - nb; j++)
            if (hs(j, j + nb - 1) == hb(0, nb - 1))
                vj.push_back(j);
        vector<int> res;
        for (auto i: vi) {
            auto l = lower_bound(vj.begin(), vj.end(), i - k);
            auto r = lower_bound(vj.begin(), vj.end(), i + k + 1);
            if (l != r)
                res.push_back(i);
        }
        return res;
    }

    class shash {//字符串哈希模板
    public:
        using ll = long long;
        vector<ll> pres;
        vector<ll> epow;
        ll e, p;

        shash(string &s, ll e, ll p) {
            int n = s.size();
            this->e = e;
            this->p = p;
            pres = vector<ll>(n + 1);
            epow = vector<ll>(n + 1);
            epow[0] = 1;
            for (int i = 0; i < n; i++) {
                pres[i + 1] = (pres[i] * e + s[i]) % p;
                epow[i + 1] = (epow[i] * e) % p;
            }
        }

        ll operator()(int l, int r) {
            ll res = (pres[r + 1] - pres[l] * epow[r - l + 1] % p) % p;
            return (res + p) % p;
        }
    };
};