暴力: 排序后遍历判断

class Solution {
public:bool isGood(vector<int> &nums) {sort(nums.begin(), nums.end());for (int i = 0; i < nums.size() - 1; i++)if (nums[i] != i + 1)return false;return nums.back() == nums.size() - 1;}
};

模拟: 单独对元音字母排序并对元音字母所在的位置按序赋值

class Solution {
public:string sortVowels(string s) {set<char> se{'a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'};vector<char> li;for (auto c: s)if (se.count(c))li.push_back(c);sort(li.begin(), li.end());int i = 0;for (auto &c: s)if (se.count(c))c = li[i++];return s;}
};

动态规划: 定义$p_{i,0}$为在$nums[0,i]$上最后一个访问的元素为偶数的最大得分和, $p_{i,1}$为在$nums[0,i]$上最后一个访问的元素为奇数的最大得分和. $p_{i,j}$的状态只与$p_{i-1,0}$和$p_{i-1,1}$有关.

class Solution {
public:typedef long long ll;long long maxScore(vector<int> &nums, int x) {int n = nums.size();ll p[n][2];p[0][nums[0] & 1] = nums[0];p[0][nums[0] & 1 ^ 1] = INT64_MIN;//无效标志ll res = p[0][nums[0] & 1];for (int i = 1; i < n; i++) {p[i][0] = INT64_MIN;p[i][1] = INT64_MIN;if (p[i - 1][nums[i] & 1] != INT64_MIN)p[i][nums[i] & 1] = max(p[i - 1][nums[i] & 1], p[i - 1][nums[i] & 1] + nums[i]);if (p[i - 1][nums[i] & 1 ^ 1]!= INT64_MIN)p[i][nums[i] & 1] = max(p[i][nums[i] & 1], p[i - 1][nums[i] & 1 ^ 1] + nums[i] - x);p[i][nums[i] & 1 ^ 1] = p[i - 1][nums[i] & 1 ^ 1];res = max({res, p[i][0], p[i][1]});}return res;}
};

动态规划: 定义$p_{v,mx}$为将 $v$ 表示成一些互不相同且其中最大值为$mx$的正整数的 x 次幂之和的方案数. 有状态转移方程: $p_{v,mx}={\sum }_{k=0}^{mx-1}{p}_{v-m{x}^x,k}$, 实现时可以用前缀和减少计算: $s_{v,i}={\sum }_{k=0}^i{p}_{v,k}$

typedef long long ll;ll fpow(ll x, ll n) {//快速幂: x^nll res = 1;for (ll cur = 1, e = x; n; e = e * e, n >>= 1)if (n & 1)res *= e;return res;
}class Solution {
public:ll mod = 1e9 + 7;int numberOfWays(int n, int x) {ll p[n + 1][n + 1];ll s[n + 1][n + 1];memset(p, 0, sizeof(p));memset(s, 0, sizeof(s));p[0][0] = 1;for (int i = 0; i <= n; i++)s[0][i] = 1;for (int v = 1; v <= n; v++) {for (int mx = 1; mx <= v && v - fpow(mx, x) >= 0; mx++)p[v][mx] = s[v - fpow(mx, x)][mx - 1];for (int i = 1; i <= n; i++)s[v][i] = (s[v][i - 1] + p[v][i]) % mod;}return (s[n][n] + mod) % mod;}
};