枚举 $x$

class Solution {
public:int countSymmetricIntegers(int low, int high) {int res = 0;for (int i = low; i <= high; i++) {string s = to_string(i);if (s.size() & 1)continue;int s1 = 0, s2 = 0;for (int k = 0; k < s.size(); k++)if (k < s.size() / 2)s1 += s[k] - '0';elses2 += s[k] - '0';if (s1 == s2)res++;}return res;}
};

双指针：则若字符串操作完后为 $0$ ，设字符串长为 $n$ ，则需要$n$ 或 $n-1$ （字符串中含 0）操作使得字符串变为 $0$ ， 若字符串操作完后至少有两位数字，则其最后两位只能是 { 25 , 50 , 75 , 00 } \{25, 50, 75, 00\} {25,50,75,00} 其中之一，枚举可能的后两位，用双指针计算要得到当前枚举值的最少操作数

class Solution {
public:int minimumOperations(string num) {vector<string> tar{"25", "50", "75", "00"};int n = num.size();int res = num.find('0') == num.npos ? n : n - 1;for (auto &s: tar) {int i = s.size() - 1;int j = n - 1;int cur = 0;//得到当前枚举值的最少操作数for (; i >= 0 && j >= 0;) {if (s[i] == num[j]) {i--;j--;} else {j--;cur++;}}if (i < 0)res = min(res, cur);}return res;}
};

前缀和：设数组 $li$ 有：$l{i}_i=\{\begin{array}{cc}1& ,nums[i]\%mod=k\\ 0& ,nums[i]\%mod\ne k\end{array}$，设 $li$ 上的前缀和为 $p{s}_i=({\sum }_{j=0}^{j<i}l{i}_i)\%mod$ ，设子数组 $nums[l,r]$ 为趣味子数组，则有：$(p{s}_{r+1}-p{s}_l)\%mod=k$，即有 $p{s}_l=((p{s}_{r+1}-k)\%mod+mod)\%mod$。

class Solution {
public:using ll = long long;long long countInterestingSubarrays(vector<int> &nums, int modulo, int k) {unordered_map<int, ll> cnt;//cnt[val]: 前缀和val出现的次数cnt[0] = 1;//前缀为空int s = 0;//当前前缀和ll res = 0;for (int i = 0; i < nums.size(); i++) {if (nums[i] % modulo == k)s = (s + 1) % modulo;int s_l = ((s - k) % modulo + modulo) % modulo;res += cnt[s_l];cnt[s]++;}return res;}
};

倍增+枚举：1）预处理：设 $0$ 为树的根节点，枚举边的权重 $w\_id$，从树根开始 $dfs$ ，计算各节点 $u$ 到树根的路径上的边数 $level[u]$，以及节点 $u$ 到树根的路径上边权重为 $w\_id$ 的边的数目 $s[u][w\_id]$，求倍增数组 $p$： $p[u][j]$为与 $u$ 距离为 $2^j$的祖先节点。2）对一个查询 $(a,b)$，用倍增的方式求 $a$ 和 $b$ 的最近公共祖先 $c$ ，然后枚举 $w\_id$ ，将 $a$ 和 $b$ 间路径上的边的边权统一为 $w\_id$ 的操作数为：$$\left(level[a]-level[c]-(s[a][w\_id]-s[c][w\_id])\right)+\left(level[b]-level[c]-(s[b][w\_id]-s[c][w\_id])\right)$$

class Solution {
public:vector<int> minOperationsQueries(int n, vector<vector<int>> &edges, vector<vector<int>> &queries) {vector<pair<int, int>> e[n];//邻接表int mx_w = 0, mn_w = INT32_MAX;//最大权重、最小权重for (auto &ei: edges) {e[ei[0]].emplace_back(ei[1], ei[2]);e[ei[1]].emplace_back(ei[0], ei[2]);mx_w = max(mx_w, ei[2]);mn_w = min(mn_w, ei[2]);}int level[n], s[n][27];int p[n][15];function<void(int, int, int, int, int)> dfs = [&](int cur, int par, int lev, int sum, int w_id) {if (w_id == mn_w)//倍增数组一轮dfs即可计算for (int i = 0; i < 15; i++)p[cur][i] = i != 0 ? p[p[cur][i - 1]][i - 1] : par;level[cur] = lev;s[cur][w_id] = sum;for (auto &[j, w]: e[cur])if (j != par)dfs(j, cur, lev + 1, w == w_id ? sum + 1 : sum, w_id);};for (int i = mn_w; i <= mx_w; i++)//枚举w_iddfs(0, 0, 0, 0, i);vector<int> res;res.reserve(queries.size());for (auto &qi: queries) {int a = qi[0], b = qi[1];if (a == b) {res.push_back(0);continue;}if (level[a] < level[b])swap(a, b);int c = a;//c最终为a和b的最近公共祖先for (int step = level[a] - level[b], ind = 0; step >= (1 << ind); ind++)if (step >> ind & 1)c = p[c][ind];if (c != b) {int b_ = b;for (int ind = 14; ind >= 0; ind--) {if (p[c][ind] != p[b_][ind]) {c = p[c][ind];b_ = p[b_][ind];}}c = p[c][0];}int res_i = INT32_MAX;for (int w_id = mn_w; w_id <= mx_w; w_id++) {//枚举w_idint t1 = level[a] - level[c] - (s[a][w_id] - s[c][w_id]);int t2 = level[b] - level[c] - (s[b][w_id] - s[c][w_id]);res_i = min(res_i, t1 + t2);}res.push_back(res_i);}return res;}
};