A leetcode.cn/problems/subarrays-distinct-element-sum-of-squares-i">子数组不同元素数目的平方和 I

枚举：枚举子数组，用集合记录当前子数组中不同元素的个数

class Solution {
public:
    using ll = long long;

    int sumCounts(vector<int> &nums) {
        ll mod = 1e9 + 7;
        int n = nums.size();
        unordered_set<int> s;
        ll res = 0;
        for (int l = 0; l < n; l++) {
            for (int r = l; r < n; r++) {//子数组nums[l,r]
                s.insert(nums[r]);
                res += s.size() * s.size();
                res %= mod;
            }
            s.clear();
        }
        return (res + mod) % mod;
    }
};

---

B leetcode.cn/problems/minimum-number-of-changes-to-make-binary-string-beautiful">使二进制字符串变美丽的最少修改次数

动态规划：设 $p[i][j]$ 为将 $s[0,i-1]$ 修改为末尾为 $j$ 的美丽字符串的最少修改次数，$p[i][j]$ 可由 $p[i-2][0]$ 和 $p[i-2][1]$ 转移得到

class Solution {
public:
    int minChanges(string s) {
        int n = s.size();
        int p[n + 1][2];
        p[0][0] = 0;
        p[0][1] = 0;
        for (int i = 2; i <= n; i += 2) {
            p[i][0] = min(p[i - 2][0], p[i - 2][1]) + (s[i - 2] == '0' ? 0 : 1) + (s[i - 1] == '0' ? 0 : 1);
            p[i][1] = min(p[i - 2][0], p[i - 2][1]) + (s[i - 2] == '1' ? 0 : 1) + (s[i - 1] == '1' ? 0 : 1);
        }
        return min(p[n][0], p[n][1]);
    }
};

---

C leetcode.cn/problems/length-of-the-longest-subsequence-that-sums-to-target">和为目标值的最长子序列的长度

动态规划：设 $p[i][j]$ 为 $nums[0,i-1]$ 的和为 $j$ 的子序列的最长长度，有状态转移方程： p [ i + 1 ] [ j ] = m i n { p [ i ] [ j ] , p [ i ] [ j − n u m s [ i ] ] + 1 } p[i+1][j]=min\{p[i][j], p[i][j-nums[i]]+1 \} p[i+1][j]=min{p[i][j],p[i][j−nums[i]]+1}

class Solution {
public:
    int lengthOfLongestSubsequence(vector<int> &nums, int target) {
        int n = nums.size();
        int p[n + 1][target + 1];
        for (int i = 0; i <= n; i++)
            for (int j = 0; j <= target; j++)
                p[i][j] = -1;//初始化为无效状态
        p[0][0] = 0;//空序列

        for (int i = 0; i < n; i++) {
            for (int j = 0; j <= target; j++) {
                p[i + 1][j] = p[i][j];
                if (j - nums[i] >= 0 && p[i][j - nums[i]] != -1)
                    p[i + 1][j] = max(p[i + 1][j], p[i][j - nums[i]] + 1);
            }
        }
        return p[n][target];
    }
};

---

D leetcode.cn/problems/subarrays-distinct-element-sum-of-squares-ii">子数组不同元素数目的平方和 II

线段树：遍历数组 $nums[i]$ ，设 $c[j]$ 为子数组 $nums[j,i]$ 中不同元素数目，考虑子数组右端点从 $i-1$ 移动到 $i$ 时 $c$ 数组的变化：

1. 若 $nums[i]$ 为该元素第一次出现，则 $c$ 数组在 $[0,i]$ 区间的元素 $+1$
2. 若 $nums[i]$ 不是该元素第一次出现，设最近的一次出现为位置为 $left[i]-1$，则 $c$ 数组在 $[left[i],i]$ 区间的元素 $+1$

可以通过实现支持 区间加 和 查询区间平方和 的线段树来求解，因为 $(a_l+d{)}^2+\cdots +(a_r+d{)}^2=(a_l^2+\cdots +a_r^2)+2\times d\times (a_l+\cdots +a_r)+d^2\times (r-l+1)$，所以线段树内部可以通过维护区间和和区间平方和来实现。

class SegmentTree {
public:
    typedef long long ll;

    inline void push_down(ll index) {
        st[index << 1].lazy = 1;
        st[index << 1 | 1].lazy = 1;
        st[index << 1].mark += st[index].mark;
        st[index << 1 | 1].mark += st[index].mark;

        st[index << 1].s2 += st[index << 1].s * st[index].mark * 2 + (st[index << 1].tr - st[index << 1].tl + 1) * st[index].mark * st[index].mark;
        st[index << 1].s += st[index].mark * (st[index << 1].tr - st[index << 1].tl + 1);
        st[index << 1 | 1].s2 += st[index << 1 | 1].s * st[index].mark * 2 + (st[index << 1 | 1].tr - st[index << 1 | 1].tl + 1) * st[index].mark * st[index].mark;
        st[index << 1 | 1].s += st[index].mark * (st[index << 1 | 1].tr - st[index << 1 | 1].tl + 1);

        st[index].lazy = 0;
        st[index].mark = 0;
    }

    inline void push_up(ll index) {
        st[index].s = st[index << 1].s + st[index << 1 | 1].s;
        st[index].s2 = st[index << 1].s2 + st[index << 1 | 1].s2;
    }

    SegmentTree(int n) {
        st = vector<SegmentTreeNode>(n * 4 + 10);
        build(1, n);
    }

    void build(ll l, ll r, ll index = 1) {
        st[index].tl = l;
        st[index].tr = r;
        st[index].lazy = 0;
        st[index].mark = 0;
        if (l == r) {
            st[index].s = 0;
            st[index].s2 = 0;
        } else {
            ll mid = (l + r) >> 1;
            build(l, mid, index << 1);
            build(mid + 1, r, index << 1 | 1);
            push_up(index);
        }
    }

    void add(ll l, ll r, ll d, ll index = 1) {
        if (l > st[index].tr or r < st[index].tl)
            return;
        else if (l <= st[index].tl and st[index].tr <= r) {
            st[index].s2 += st[index].s * d * 2 + (st[index].tr - st[index].tl + 1) * d;
            st[index].s += (st[index].tr - st[index].tl + 1) * d;
            st[index].mark += d;
            st[index].lazy = 1;
        } else {
            if (st[index].lazy)
                push_down(index);
            add(l, r, d, index << 1);
            add(l, r, d, index << 1 | 1);
            push_up(index);
        }
    }

    ll query(ll l, ll r, ll index = 1) {
        if (l <= st[index].tl and st[index].tr <= r) {
            return st[index].s2;
        } else {
            if (st[index].lazy)
                push_down(index);
            if (r <= st[index << 1].tr)
                return query(l, r, index << 1);
            else if (l > st[index << 1].tr)
                return query(l, r, index << 1 | 1);
            return query(l, r, index << 1) + query(l, r, index << 1 | 1);
        }
    }

private:
    struct SegmentTreeNode {
        ll tl;
        ll tr;
        ll s;//区间和
        ll s2;//区间平方和
        ll mark;
        int lazy;
    };
    vector<SegmentTreeNode> st;
};
class Solution {
public:
    using ll = long long;
    ll mod = 1e9 + 7;

    int sumCounts(vector<int> &nums) {
        int n = nums.size();
        unordered_map<int, int> vis;
        vector<int> left(n);
        for (int i = 0; i < n; i++) {
            if (vis.count(nums[i]))
                left[i] = vis[nums[i]] + 1;
            else
                left[i] = 0;
            vis[nums[i]] = i;
        }
        SegmentTree st(n);
        ll res = 0;
        for (int i = 0; i < n; i++) {
            st.add(left[i] + 1, i + 1, 1);
            res = (res + st.query(1, i + 1)) % mod;
        }
        return (res + mod) % mod;
    }
};