模拟

class Solution {
public:int returnToBoundaryCount(vector<int> &nums) {int s = 0;int res = 0;for (auto x: nums) {s += x;if (s == 0)res++;}return res;}
};

枚举：若经过 $i$ 秒后 $word$ 可以恢复到其初始状态，则说明 $word$ 长为 $n-i\times k$的后缀与长为 $n-i\times k$的前缀相等，$n$ 为 $word$ 的长度，升序枚举 $i$ 直到满足条件

class Solution {
public:int minimumTimeToInitialState(string word, int k) {int n = word.size();int i = 1;for (; i * k < n; i++) {int tag = 1;for (int j = i * k; j < n; j++)if (word[j] != word[j - i * k])tag = 0;if (tag)return i;}return i;}
};

模拟：枚举网格中的 $3\times 3$ 的子网格，判断是否是区域，同时记录各位置所属于的区域数和所属于的区域的平均强度之和，最后求网格的区域平均强度

class Solution {
public:vector<vector<int>> resultGrid(vector<vector<int>> &image, int threshold) {int m = image.size(), n = image[0].size();vector<vector<int>> tot(m, vector<int>(n));vector<vector<int>> cnt(m, vector<int>(n));for (int i = 0; i + 2 < m; i++)for (int j = 0; j + 2 < n; j++) {int tag = 1;//判断是否是区域for (int x = i; x < i + 3; x++)for (int y = j + 1; y < j + 3; y++)if (abs(image[x][y] - image[x][y - 1]) > threshold)tag = 0;for (int y = j; y < j + 3; y++)for (int x = i + 1; x < i + 3; x++)if (abs(image[x][y] - image[x - 1][y]) > threshold)tag = 0;if (tag) {int s = 0;for (int x = i; x < i + 3; x++)for (int y = j; y < j + 3; y++)s += image[x][y];s /= 9;//当前区域的平均强度for (int x = i; x < i + 3; x++)for (int y = j; y < j + 3; y++) {cnt[x][y]++;//所属区域数目+1tot[x][y] += s;//所属区域的平均强度之和+s}}}for (int i = 0; i < m; i++)for (int j = 0; j < n; j++)if (!cnt[i][j])tot[i][j] = image[i][j];elsetot[i][j] /= cnt[i][j];return tot;}
};

字符串哈希 + 枚举：若经过 $i$ 秒后 $word$ 可以恢复到其初始状态，则说明 $word$ 长为 $n-i\times k$的后缀与长为 $n-i\times k$的前缀相等，$n$ 为 $word$ 的长度，升序枚举 $i$ 直到满足条件

class Solution {
public:int minimumTimeToInitialState(string word, int k) {int n = word.size();int i = 1;srand(time(0));shash h(word, 2333 + rand() % 100, 1e9 + rand() % 100);for (; i * k < n; i++) {     if (h(i * k, n - 1) == h(0, n - 1 - i * k))//s[i*k,n-1]和s[0,n-1-i*k]是否相等return i;}return i;}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;}};
};