传送门

6354. 找出数组的串联值

题目

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思路

前后指针

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代码

class Solution {
public:long long findTheArrayConcVal(vector<int>& nums) {long long res = 0;int n = nums.size();for (int i = 0, j = n - 1; i < j; i++, j--) {res += stol(to_string(nums[i]) + to_string(nums[j]));}if (n & 1) res += nums[n / 2];return res; }
};

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6355. 统计公平数对的数目

题目

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思路

$lower<=nums[i]+nums[j]<=upper$

转化：

$lower-nums[i]<=nums[j]<=upper-nums[i]$

倒序遍历，动态查找后缀的在区间如上范围内的值的个数。那么考虑用树状数组获取区间内值的个数。

由于范围过大，且做差会造成负数，树状数组下标从1开始。所以换用map的树状数组，平且值向右平移到正数。

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代码

class Solution {
public:using ll = long long;static constexpr ll N = 2e9 + 7;class BIT {public:BIT() {}unordered_map<ll, int> tr;void add(ll x, int v = 1) { for (; x < N << 1; x += x & -x) tr[x] += v; }ll sum(ll x) { ll res = 0; for (; x; x -= x & -x) res += tr[x]; return res; }}; // 切记用 BIT tr; tr.add(); tr.sum();long long countFairPairs(vector<int>& a, int L, int R) {ll res = 0;BIT tree;for (int i = a.size() - 1; i >= 0; i--) {res += tree.sum(R - a[i] + N) - tree.sum(L - a[i] - 1 + N);tree.add(a[i] + N);}return res;}
};

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6356. 子字符串异或查询

题目

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思路

映射数值对应的最小开始下标即可，数值在1e9，并且无前导 ‘0’，所以只需要 O(30n)。

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代码

class Solution {
public:vector<vector<int>> substringXorQueries(string s, vector<vector<int>>& q) {vector<vector<int>> res;map<int, int> mp;int mn = 1e9;for (int i = 0; i < s.size(); i++) {int x = 0;for (int j = 0; j < 31 && i + j < s.size(); j++) {if (s[i + j] - '0' == 0 && x == 0) {mn = min(mn, i);break;}x = x * 2 + s[i + j] - '0';if (mp.count(x)) {mp[x] = min(mp[x], i);} else {mp[x] = i;}}}for (int i = 0; i < q.size(); i++) {int x = q[i][0] ^ q[i][1];int y = x;int cnt = 0;while (y) { cnt++; y /= 2; }vector<int> v{-1, -1};if (cnt == 0 && mn != 1e9) {v[0] = mn;v[1] = mn;}if (mp.count(x)) {v[0] = mp[x];v[1] = v[0] + cnt - 1;}res.push_back(v);}return res;}
};

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6357. 最少得分子序列

题目

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思路

维护前后缀数组，$l[i],r[i]$：表示字符串 $s$ 的前 $i$ 位子序列在字符串 $t$ 中匹配的最长前缀。$r$ 同理最长后缀。

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代码

class Solution {
public:int minimumScore(string s, string t) {int n = s.size(); vector<int> l(n, 0), r(n, 0); int j = 0;for (int i = 0; i < s.size(); i++) {if (j < t.size() && s[i] == t[j]) ++j;l[i] = j;}j = t.size() - 1;for (int i = s.size() - 1; i >= 0; i--) {if (j >= 0 && s[i] == t[j]) --j;r[i] = t.size() - j - 1;}int res = 1e9; for (int i = 0; i < n - 1; i++) {int L = l[i], R = r[i + 1];if (L + R >= t.size()) return 0; // 都能匹配上，说明t本就是s的子序列res = min(res, (int)((t.size() - R - 1) - (L) + 1));}res = min(res, (int)t.size() - r[0]);res = min(res, (int)t.size() - l[n - 1]);return res;}
};

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