数据量小，直接枚举数对

class Solution {
public:int countPairs(vector<int> &nums, int target) {int n = nums.size();int res = 0;for (int i = 0; i < n; i++)for (int j = 0; j < i; j++)if (nums[i] + nums[j] < target)res++;return res;}
};

贪心：从两字符串的头部开始匹配，若两串中的字符匹配则两串中的下标都+1，否则仅str1中的下标+1，最终通过str2中的下标判断str2是否已匹配完。

class Solution {
public:bool canMakeSubsequence(string str1, string str2) {int m = str1.size(), n = str2.size();int i = 0, j = 0;for (; i < m && j<n;) {if (str1[i] == str2[j] || ('a' + (str1[i] - 'a' + 1) % 26) == str2[j]) {i++;j++;} else {i++;}}return j == n;}
};

脑筋急转弯：设$n$为数组长度，设原数组$nums$上构建的新数组$res$中最长上升子序列的长度为$l$，则$n-l$即为答案。数据范围小可以直接$O(n^2)$动态规划求最长上升子序列，设$p[i]$为以$res$中以$res[i]$结尾的最长上升子序列的长度，有状态转移方程： p [ i ] = m a x { 1 m a x { p [ j ] + 1 ∣ 0 ≤ j < i , r e s [ j ] < r e s [ i ] } p[i]=max\left\{\begin{matrix} 1 \\ max\{p[j]+1 \;|\; 0\le j<i,res[j]<res[i]\} \end{matrix}\right. p[i]=max{1max{p[j]+1∣0≤j<i,res[j]<res[i]}​

class Solution {
public:int minimumOperations(vector<int> &nums) {int n = nums.size();vector<int> li[4];for (int i = 0; i < n; i++)li[nums[i]].push_back(i);vector<int> vec;//nums构建的res数组for (int i = 1; i <= 3; i++)if (!li[i].empty())for (auto x: li[i])vec.push_back(x);int p[n];for (int i = 0; i < n; i++) {p[i] = 1;for (int j = 0; j < i; j++)if (vec[j] < vec[i])p[i] = max(p[i], p[j] + 1);}return n - *max_element(p, p + n);}
};

数位$dp$：设$comp(x,k)$为满足$0\le i\le x$的美丽的$i$的个数，则原问题答案可表示为$comp(high,k)-comp(low-1,k)$。使用数位$dp$求$comp(x,k)$：设$p[loc][val][eq][oe][mod]$为满足 “当前数位下标为$loc$、当前数位之前是否有数字（有：$val=1$，无：$val=0$）、当前数位之前是否严格等于$x$（是：$eq=1$，否：$eq=0$）、当前奇偶情况$oe$（遇到奇数位+1，遇到偶数维-1），当前数位之前的数对$k$取模的余数为$mod$ ”的情况下的满足$0\le i\le x$的美丽的$i$的个数，分类讨论实现状态转移。

class Solution {
public:int comp(int x, int k) {string s = to_string(x);int n = s.size();int p[n + 1][2][2][2 * n + 1][k];memset(p, -1, sizeof(p));int c = n;//oe维度需要的偏移量，防止下标出现负数function<int(int, int, int, int, int)> get = [&](int loc, int val, int eq, int oe, int mod) {if (p[loc][val][eq][oe][mod] != -1)return p[loc][val][eq][oe][mod];int &cur = p[loc][val][eq][oe][mod];if (loc == n)//已遍历完所有数位return cur = (val && oe == c && mod == 0) ? 1 : 0;cur = 0;if (!val) {//当前数位之前无数字cur += get(loc + 1, 0, 0, oe, mod);//当前数位继续无数字if (eq)//当前数位之前严格等于xfor (int i = 1; i <= s[loc] - '0'; i++)cur += get(loc + 1, 1, i != s[loc] - '0' ? 0 : 1, oe + (i % 2 == 1 ? 1 : -1), (mod * 10 + i) % k);else//当前数位之前不严格等于xfor (int i = 1; i <= 9; i++)cur += get(loc + 1, 1, 0, oe + (i % 2 == 1 ? 1 : -1), (mod * 10 + i) % k);} else {当前数位之前有数字if (eq)//当前数位之前严格等于xfor (int i = 0; i <= s[loc] - '0'; i++)cur += get(loc + 1, 1, i != s[loc] - '0' ? 0 : 1, oe + (i % 2 == 1 ? 1 : -1), (mod * 10 + i) % k);else//当前数位之前不严格等于xfor (int i = 0; i <= 9; i++)cur += get(loc + 1, 1, 0, oe + (i % 2 == 1 ? 1 : -1), (mod * 10 + i) % k);}return cur;};return get(0, 0, 1, c + 0, 0);}int numberOfBeautifulIntegers(int low, int high, int k) {int a = comp(high, k), b = comp(low - 1, k);return a - b;}
};