Q1. 检测相邻递增子数组 I
原题链接
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
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 FalseQ2. 检测相邻递增子数组 II
原题链接
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
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 loQ3. 好子序列的元素之和
原题链接
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
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
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
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;
}
};