LeetCode 2024/8 每日一题合集
2024-7-1 LCP 40. 心算挑战
代码实现
class Solution {
public:
int maxmiumScore(vector<int>& cards, int cnt) {
int n = size(cards);
std::sort(cards.rbegin(), cards.rend());
int sum = std::accumulate(cards.begin(), cards.begin() + cnt, 0);
int odd = -1, even = -1;
for (int i = 0; i < cnt; ++i) {
if (cards[i] % 2) {
odd = i;
} else {
even = i;
}
}
int ans = 0;
if (cnt == n || sum % 2 == 0) {
ans = sum;
} else {
{
int now = -1;
for (int i = cnt; i < n; ++i) {
if (cards[i] % 2 == 0) {
now = i;
break;
}
}
if (now == -1) {
ans = std::max(ans, sum - cards[odd]);
} else {
ans = std::max(ans, sum - cards[odd] + cards[now]);
}
}
{
int now = -1;
for (int i = cnt; i < n; ++i) {
if (cards[i] % 2) {
now = i;
break;
}
}
if (now == -1) {
ans = std::max(ans, sum - cards[odd]);
} else if (even != -1) {
ans = std::max(ans, sum - cards[even] + cards[now]);
}
}
}
return ans % 2 == 0 ? ans : 0;
}
};
2024-8-2 3128. 直角三角形
代码实现
class Solution {
public:
long long numberOfRightTriangles(vector<vector<int>>& grid) {
int n = size(grid), m = size(grid[0]);
std::vector<std::vector<int>> p1(n + 1, std::vector<int>(m + 1));
std::vector<std::vector<int>> p2(n + 1, std::vector<int>(m + 1));
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
p1[i + 1][j + 1] = p1[i + 1][j] + grid[i][j];
}
}
for (int j = 0; j < m; ++j) {
for (int i = 0; i < n; ++i) {
p2[i + 1][j + 1] = p2[i][j + 1] + grid[i][j];
}
}
using i64 = long long;
i64 ans = 0;
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= m; ++j) if (grid[i - 1][j - 1]) {
int left = p1[i][j - 1] - p1[i][0], right = p1[i][m] - p1[i][j];
int up = p2[i - 1][j] - p2[0][j], down = p2[n][j] - p2[i][j];
ans += up * right + right * down + down * left + left * up;
}
}
return ans;
}
};
2024-8-3 3143. 正方形中的最多点数
代码实现
class Solution {
public:
int maxPointsInsideSquare(vector<vector<int>>& points, string s) {
int n = size(points), ans = 0;
auto check = [&](int size) {
int vis = 0;
for (int i = 0; i < n; ++i) {
if (abs(points[i][0]) <= size && abs(points[i][1]) <= size) {
char c = s[i] - 'a';
if (vis >> c & 1) {
return false;
}
vis |= 1 << c;
}
}
ans = __builtin_popcount(vis);
return true;
};
int l = 0, r = 1e9 + 1;
while (l < r) {
int mid = l + r >> 1;
if (check(mid)) {
l = mid + 1;
} else {
r = mid;
}
}
return ans;
}
};
2024-8-4 572. 另一棵树的子树
补签，但是连胜断了QWQ
代码实现
class Solution {
public:
bool isSubtree(TreeNode* A, TreeNode* B) {
if (B == NULL) return false;
auto check = [&](auto &&self, TreeNode *a, TreeNode *b)->bool {
if (a == nullptr && b == nullptr) return true;
if (b == nullptr) return false;
if (a == nullptr) return false;
if (a->val != b->val) return false;
return self(self, a->left, b->left) && self(self, a->right, b->right);
};
auto dfs = [&](auto &&self, TreeNode *root)->bool {
if (root == NULL) return false;
if (root->val == B->val) {
if (check(check, root, B)) return true;
}
if (self(self, root->right)) return true;
if (self(self, root->left)) return true;
return false;
};
return dfs(dfs, A);
}
};
*2024-8-5 600. 不含连续1的非负整数
代码实现
class Solution {
public:
int findIntegers(int n) {
std::vector<int> f(31, 1);
for (int i = 2; i < 31; ++i) {
f[i] = f[i - 1] + f[i - 2];
}
int pre = 0, ans = 0;
for (int i = 29; i >= 0; --i) {
if ((n & (1 << i))) {
ans += f[i + 1];
if (pre == 1) break;
pre = 1;
} else pre = 0;
ans += i == 0;
}
return ans;
}
};
*2024-8-6 3129. 找出所有稳定的二进制数组 I
代码实现
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;
}
template <int P>
struct MInt {
int x;
constexpr MInt() : x{} {}
constexpr MInt(i64 x) : x{norm(x % P)} {}
constexpr int norm(int x) const {
if (x < 0) x += P;
if (x >= P) x -= P;
return x;
}
constexpr int val() const { return x; }
explicit constexpr operator int() const { return x; }
constexpr MInt operator-() const {
MInt res;
res.x = norm(P - x);
return res;
}
constexpr MInt inv() const {
assert(x != 0);
return power(*this, P - 2);
}
constexpr MInt &operator*=(MInt rhs) {
x = 1ll * x * rhs.x % P;
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 V, int P>
constexpr MInt<P> CInv = MInt<P>(V).inv();
// constexpr int P = 998244353;
constexpr int P = 1e9 + 7;
using Z = MInt<P>;
struct Comb {
int n;
std::vector<Z> _fac, _invfac, _inv;
Comb() : n{0}, _fac{1}, _invfac{1}, _inv{0} {}
Comb(int n) : Comb() { init(n); }
void init(int m) {
if (m <= n) return;
_fac.resize(m + 1), _invfac.resize(m + 1), _inv.resize(m + 1);
for (int i = n + 1; i <= m; ++ i) _fac[i] = _fac[i - 1] * i;
_invfac[m] = _fac[m].inv();
for (int i = m; i > n; -- i) _invfac[i - 1] = _invfac[i] * i, _inv[i] = _invfac[i] * _fac[i - 1];
n = m;
}
Z fac(int m) {
if (m > n) init(2 * m);
return _fac[m];
}
Z invfac(int m) {
if (m > n) init(2 * m);
return _invfac[m];
}
Z inv(int m) {
if (m > n) init(2 * m);
return _inv[m];
}
Z binom(int n, int m) {
if (n < m || m < 0) return 0;
return fac(n) * invfac(m) * invfac(n - m);
}
} comb;
class Solution {
public:
int numberOfStableArrays(int zero, int one, int limit) {
int n = zero + one;
if (zero > one) {
std::swap(zero, one);
}
std::vector<Z> f0(zero + 3);
for (int i = (zero - 1) / limit + 1; i <= zero; ++i) {
f0[i] = comb.binom(zero - 1, i - 1);
for (int j = 1; j <= (zero - i) / limit; ++j) {
f0[i] += (j % 2 ? -1 : 1) * comb.binom(i, j) * comb.binom(zero - j * limit - 1, i - 1);
}
}
std::vector<Z> f1(one + 3);
for (int i = (one - 1) / limit + 1; i <= one; ++i) {
f1[i] = comb.binom(one - 1, i - 1);
for (int j = 1; j <= (one - i) / limit; ++j) {
f1[i] += (j % 2 ? -1 : 1) * comb.binom(i, j) * comb.binom(one - j * limit - 1, i - 1);
}
}
Z ans = 0;
for (int i = (one - 1) / limit + 1; i <= std::min(zero + 1, one); ++i) {
ans += (f0[i - 1] + 2 * f0[i] + f0[i + 1]) * f1[i];
}
return (int)ans;
}
};
*2024-8-7 3130. 找出所有稳定的二进制数组 II
代码实现
同前一天
*2024-8-8 3131. 找出与数组相加的整数 I
代码实现
class Solution {
public:
int addedInteger(vector<int>& nums1, vector<int>& nums2) {
return std::ranges::max(nums2) - std::ranges::max(nums1);
}
};
*2024-8-9 3132. 找出与数组相加的整数 II
代码实现
class Solution {
public:
int minimumAddedInteger(vector<int>& nums1, vector<int>& nums2) {
std::ranges::sort(nums1);
std::ranges::sort(nums2);
for (int i = 2; i > 0; --i) {
int x = nums2[0] - nums1[i];
int j = 0;
for (int k = i; k < size(nums1); k++) {
if (nums2[j] == nums1[k] + x && ++j == size(nums2)) {
return x;
}
}
}
return nums2[0] - nums1[0];
}
};
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