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;
}
};
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;
}
};
};
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 m i d mid mid 的非负数（ 0 0 0 的价值和为 0 0 0 不影响答案）的价值和，枚举二进制中的满足 i % x = = 0 i \% x == 0 i%x==0 的第 i i i 位， 0 ∼ m i d 0\sim mid 0∼mid 的第 i i i 位形成长为 2 i 2^i 2i 的循环数列的前 m i d + 1 mid+1 mid+1 项： 0 , ⋯ , 0 ⏟ 2 i − 1 个 0 , 1 , ⋯ , 1 ⏟ 2 i − 1 个 1 ⋯ \underset{2^{i-1}个0}{\underbrace{0,\cdots,0} } ,\underset{2^{i-1}个1}{\underbrace{1,\cdots,1} } \cdots 2i−1个0 0,⋯,0,2i−1个1 1,⋯,1⋯，所以可以直接求 0 ∼ m i d 0\sim mid 0∼mid 中第 i i i 位为 1 1 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;
}
};
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 ≤ i ≤ s . l e n g t h − a . l e n g t h 0 \le i \le s.length - a.length 0≤i≤s.length−a.length 和 s [ i , ( i + a . l e n g t h − 1 ) ] = a s[i,(i + a.length - 1)]=a s[i,(i+a.length−1)]=a 的 i i i 构成的有序数组 v i vi vi，和满足 0 ≤ j ≤ s . l e n g t h − b . l e n g t h 0 \le j \le s.length - b.length 0≤j≤s.length−b.length 和 s [ j , ( j + b . l e n g t h − 1 ) ] = b s[j,(j + b.length - 1)]=b s[j,(j+b.length−1)]=b 的 j j j 构成的有序数组 v j vj vj ，然后枚举 v i vi vi 中的 i i i ，在 v j vj vj 中二分查找 j ∈ [ i − k , i + k − 1 ) j\in [i-k,i+k-1) j∈[i−k,i+k−1) ，如果存在这样的 j j j ，则将 i i 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;
}
};
};
