LeetCode 第 423 场周赛个人题解

Q1. 检测相邻递增子数组 I

原题链接

Q1. 检测相邻递增子数组 I

思路分析

线性DP

定义 f[i] 为以下标 i 为结尾的最长递增子数组长度

f[i] = f[i - 1] + 1，如果 nums[i] > nums[i - 1]

然后看一下是否存在 f[i] >= k and f[i + k] >= k

时间复杂度：O(N)

AC代码

python 复制代码

class Solution:
    def hasIncreasingSubarrays(self, nums: List[int], k: int) -> bool:
        n = len(nums)
        f = [1] * n
        for i in range(1, n):
            if nums[i] > nums[i - 1]:
                f[i] = f[i - 1] + 1
        for i in range(n - k):
            if f[i] >= k and f[i + k] >= k:
                return True
        return False

Q2. 检测相邻递增子数组 II

原题链接

Q2. 检测相邻递增子数组 II

思路分析

二分 + T1

直接二分，把T1拿过来check

时间复杂度：O(nlogn)

AC代码

python 复制代码

class Solution:
    def check(self, nums: List[int], k: int) -> bool:
        n = len(nums)
        f = [1] * n
        for i in range(1, n):
            if nums[i] > nums[i - 1]:
                f[i] = f[i - 1] + 1
        for i in range(n - k):
            if f[i] >= k and f[i + k] >= k:
                return True
        return False
    def maxIncreasingSubarrays(self, nums: List[int]) -> int:
        n = len(nums)
        lo = 0
        hi = n
        while lo < hi:
            x = (lo + hi + 1) // 2
            if self.check(nums, x):
                lo = x
            else:
                hi = x - 1
        return lo

Q3. 好子序列的元素之和

原题链接

Q3. 好子序列的元素之和

思路分析

dp + 计数

注意到值域很小，直接dp计数

f[x] 为 x 结尾的好序列的和，g[x] 为 x 结尾的好序列的个数

递推关系显然

时间复杂度：O(M)

AC代码

cpp 复制代码

using i64 = long long;

template<class T>
constexpr T power(T a, i64 b) {
    T res = 1;
    for (; b; b /= 2, a *= a) {
        if (b % 2) {
            res *= a;
        }
    }
    return res;
}

constexpr i64 mul(i64 a, i64 b, i64 p) {
    i64 res = a * b - i64(1.L * a * b / p) * p;
    res %= p;
    if (res < 0) {
        res += p;
    }
    return res;
}
template<i64 P>
struct MLong {
    i64 x;
    constexpr MLong() : x{} {}
    constexpr MLong(i64 x) : x{norm(x % getMod())} {}
    
    static i64 Mod;
    constexpr static i64 getMod() {
        if (P > 0) {
            return P;
        } else {
            return Mod;
        }
    }
    constexpr static void setMod(i64 Mod_) {
        Mod = Mod_;
    }
    constexpr i64 norm(i64 x) const {
        if (x < 0) {
            x += getMod();
        }
        if (x >= getMod()) {
            x -= getMod();
        }
        return x;
    }
    constexpr i64 val() const {
        return x;
    }
    explicit constexpr operator i64() const {
        return x;
    }
    constexpr MLong operator-() const {
        MLong res;
        res.x = norm(getMod() - x);
        return res;
    }
    constexpr MLong inv() const {
        assert(x != 0);
        return power(*this, getMod() - 2);
    }
    constexpr MLong &operator*=(MLong rhs) & {
        x = mul(x, rhs.x, getMod());
        return *this;
    }
    constexpr MLong &operator+=(MLong rhs) & {
        x = norm(x + rhs.x);
        return *this;
    }
    constexpr MLong &operator-=(MLong rhs) & {
        x = norm(x - rhs.x);
        return *this;
    }
    constexpr MLong &operator/=(MLong rhs) & {
        return *this *= rhs.inv();
    }
    friend constexpr MLong operator*(MLong lhs, MLong rhs) {
        MLong res = lhs;
        res *= rhs;
        return res;
    }
    friend constexpr MLong operator+(MLong lhs, MLong rhs) {
        MLong res = lhs;
        res += rhs;
        return res;
    }
    friend constexpr MLong operator-(MLong lhs, MLong rhs) {
        MLong res = lhs;
        res -= rhs;
        return res;
    }
    friend constexpr MLong operator/(MLong lhs, MLong rhs) {
        MLong res = lhs;
        res /= rhs;
        return res;
    }
    friend constexpr std::istream &operator>>(std::istream &is, MLong &a) {
        i64 v;
        is >> v;
        a = MLong(v);
        return is;
    }
    friend constexpr std::ostream &operator<<(std::ostream &os, const MLong &a) {
        return os << a.val();
    }
    friend constexpr bool operator==(MLong lhs, MLong rhs) {
        return lhs.val() == rhs.val();
    }
    friend constexpr bool operator!=(MLong lhs, MLong rhs) {
        return lhs.val() != rhs.val();
    }
};

template<>
i64 MLong<0LL>::Mod = i64(1E18) + 9;

template<int P>
struct MInt {
    int x;
    constexpr MInt() : x{} {}
    constexpr MInt(i64 x) : x{norm(x % getMod())} {}
    
    static int Mod;
    constexpr static int getMod() {
        if (P > 0) {
            return P;
        } else {
            return Mod;
        }
    }
    constexpr static void setMod(int Mod_) {
        Mod = Mod_;
    }
    constexpr int norm(int x) const {
        if (x < 0) {
            x += getMod();
        }
        if (x >= getMod()) {
            x -= getMod();
        }
        return x;
    }
    constexpr int val() const {
        return x;
    }
    explicit constexpr operator int() const {
        return x;
    }
    constexpr MInt operator-() const {
        MInt res;
        res.x = norm(getMod() - x);
        return res;
    }
    constexpr MInt inv() const {
        assert(x != 0);
        return power(*this, getMod() - 2);
    }
    constexpr MInt &operator*=(MInt rhs) & {
        x = 1LL * x * rhs.x % getMod();
        return *this;
    }
    constexpr MInt &operator+=(MInt rhs) & {
        x = norm(x + rhs.x);
        return *this;
    }
    constexpr MInt &operator-=(MInt rhs) & {
        x = norm(x - rhs.x);
        return *this;
    }
    constexpr MInt &operator/=(MInt rhs) & {
        return *this *= rhs.inv();
    }
    friend constexpr MInt operator*(MInt lhs, MInt rhs) {
        MInt res = lhs;
        res *= rhs;
        return res;
    }
    friend constexpr MInt operator+(MInt lhs, MInt rhs) {
        MInt res = lhs;
        res += rhs;
        return res;
    }
    friend constexpr MInt operator-(MInt lhs, MInt rhs) {
        MInt res = lhs;
        res -= rhs;
        return res;
    }
    friend constexpr MInt operator/(MInt lhs, MInt rhs) {
        MInt res = lhs;
        res /= rhs;
        return res;
    }
    friend constexpr std::istream &operator>>(std::istream &is, MInt &a) {
        i64 v;
        is >> v;
        a = MInt(v);
        return is;
    }
    friend constexpr std::ostream &operator<<(std::ostream &os, const MInt &a) {
        return os << a.val();
    }
    friend constexpr bool operator==(MInt lhs, MInt rhs) {
        return lhs.val() == rhs.val();
    }
    friend constexpr bool operator!=(MInt lhs, MInt rhs) {
        return lhs.val() != rhs.val();
    }
};

template<>
int MInt<0>::Mod = 998244353;

template<int V, int P>
constexpr MInt<P> CInv = MInt<P>(V).inv();

constexpr int P = 1000000007;
using Z = MInt<P>;

class Solution {
public:
    int sumOfGoodSubsequences(vector<int>& nums) {
        const int M = std::ranges::max(nums) + 1;
        std::vector<Z> f(M + 1), g(M + 1);
        Z res = 0;
        for (int x : nums) {
            if (x > 0) {
                g[x] += g[x - 1];
                f[x] += f[x - 1] + g[x - 1] * x;
            }
            if (x + 1 <= M) {
                g[x] += g[x + 1];
                f[x] += f[x + 1] + g[x + 1] * x;
            }
            g[x] += 1;
            f[x] += x;
        }
        for (Z x : f) {
            res += x;
        }
        return res.x;
    }
};

Q4. 统计小于 N 的 K 可约简整数

原题链接

Q4. 统计小于 N 的 K 可约简整数

思路分析

数位dp

小于等于 n 的含 k 个1的数字个数如何求？------数位dp

合法的 k 怎么预处理 ------ 暴力

关于数位dp： 数位dp详解，记忆化搜索，递推，OJ精讲_数位dp记忆化搜索-CSDN博客

时间复杂度：O(N^2)

AC代码

cpp 复制代码

const int P = 1'000'000'007;

constexpr int N = 800;
int memo[N][N + 1];

class Solution {
public:
    int countKReducibleNumbers( std::string s, int k) {
        int n = s.size();
        std::vector<bool> f(n + 1, false);
        f[1] = true;
        
        for (int x = 2; x <= n; ++x) {
            int cnt = 0;
            int t = x;
            int temp_x = x;
            while (temp_x != 1 && cnt < k) {
                temp_x = __builtin_popcount(temp_x);
                cnt++;
            }
            f[t] = (cnt < k);
        }

        memset(memo, -1, sizeof(memo));
        
        auto dfs = [&](auto &&self, int n, int pre, bool lim) -> int {
            if (pre < 0) return 0;
            if (n == -1) return pre == 0 ? 1 : 0;

            if (lim && ~memo[n][pre]) return memo[n][pre];

            int top = lim ? 1 : (s[n] - '0');
            int res = 0;

            for (int x = 0; x <= top; ++x) {
                res += self(self, n - 1, pre - x, lim || x < (s[n] - '0'));
                if (res >= P) res -= P;
            }
            return memo[n][pre] = res;
        };

        std::reverse(s.begin(), s.end());
        int res = 0;

        int c1 = 0;
        for (char x : s) {
            if (x == '1') c1++;
        }
        
        for (int x = 0; x <= n; ++x) {
            if (!f[x]) continue;
            
            res += dfs(dfs, n - 1, x, false);
            if (res >= P) res -= P;
        }
        
        if (f[c1]) {
            res = (res - 1 + P) % P;
        }
        
        return res;
    }
};