0816 京东笔试答案第一题, 减少逆序对数量对于每个 ai, 找到 所有 aj = a[i] + 1 且 j < i 的 j, 则对于选择 j < L <= i 的 L 贡献 +1, 然后做一个前缀和取最大值即可代码```#include <iostream>#include <cstring>#include <algorithm>#include <vector>#include <unordered_map>using namespace std;void solve(){int n; cin >> n;vector<int> a(n + 10, 0), b(n + 10, 0), c(n + 10, 0);unordered_map<int, vector<int>> pos;for (int i = 1; i <= n; i ++){cin >> a[i];}for (int i = 1; i <= n; i ++){int v =a[i];int t = v + 1;if (pos.find(t) != pos.end()){for (int j : pos[t]){if (j < i){b[j + 1] ++;b[i + 1] --;}}}pos[v].push_back(i);}int ans = 0;int tmp = 0;for (int i = 1; i <= n; i ++){tmp += b[i];ans = max(ans, tmp);}cout << ans << endl;}int main(){int _ = 1;cin >> _;while (_ --){solve();}}```第二题, 跳水打分解法类似求区间最值, 维护每个长度为 m 的区间的最大值和最小值, 然后遍历过去即可算出答案。代码```#include <bits/stdc++.h>#define PII pair<long, long>#define x first#define y secondusing namespace std;int main(){int n, m;cin >> n >> m;priority_queue<PII> maxPq;priority_queue<PII, vector<PII>, greater<PII>> minPq;vector<long long> a(n + 10);double ans = 0;long long tmp = 0;for (int i = 1; i <= n; i ++) cin >> a[i];for (int i = 1; i <= m; i ++){maxPq.push({a[i], i});minPq.push({a[i], i});tmp += a[i];}long long minV = minPq.top().x;long long maxV = maxPq.top().x;ans = double(tmp - minV - maxV) / (m - 2);int ans2 = 1;int l = 1, r = m;while (r < n){// cout << ans << endl;tmp -= a[l]; l ++;r ++; tmp += a[r];minPq.push({a[l], l}); minPq.push({a[r], r});maxPq.push({a[l], l}); maxPq.push({a[r], r});while (minPq.top().y < l) minPq.pop();while (maxPq.top().y < l) maxPq.pop();if ((double(tmp - minPq.top().x - maxPq.top().x) / (m - 2)) > ans){ans = double(tmp - minPq.top().x - maxPq.top().x) / (m - 2);ans2 = l;}}cout << ans2 << endl;}```
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