題目傳送門
一個非常顯然的想法是記錄後面的值相鄰兩個之間在前面選了多少個數。
衆所周知(比如我就不知道,我甚至以爲它非常大),若干個和爲 $n$ 的數的乘積最大爲 $O(3^{n/3})$,最優方案是拆成若干個 3 和常數個 2。
然後 dp 即可。
時間複雜度 $O(n^23^{\frac{n + 1}{3}})$.
Code
#include <bits/stdc++.h> using namespace std; typedef bool boolean; const int N = 41; const int inf = (signed) (~0u >> 2); #define pii pair<int, long long> pii operator + (pii a, pii b) { if (a.first == b.first) return pii(a.first, a.second + b.second); return min(a, b); } int n; int p[N]; int main() { scanf("%d", &n); for (int i = 1; i <= n; i++) { scanf("%d", p + i); --p[i]; } vector<boolean> have (n, true); vector<int> lenc (n + 1, 1), lenp; vector<int> prodc (n + 2, 1), prodp; vector<pii> f {pii(0, 1)}, g; vector<int> vs; for (int i = 1; i <= n; i++) { swap(lenp, lenc); swap(prodp, prodc); swap(f, g); int id = 0; for (int j = 0; j < p[i]; j++) id += have[j]; have[p[i]] = false; lenc.clear(); prodc = {1}; int ls = -1; for (int j = 0; j < n; j++) { if (have[j]) { lenc.push_back(j - ls); prodc.push_back(prodc.back() * lenc.back()); ls = j; } } lenc.push_back(n - ls); prodc.push_back(prodc.back() * lenc.back()); f = vector<pii>(prodc.back(), pii(inf, 0)); for (int s = 0; s < prodp.back(); s++) { if (g[s].first == inf) continue; vs.clear(); int t = s; for (auto l : lenp) vs.push_back(t % l), t /= l; int dlt = 0; for (int j = id + 1; j < (signed) vs.size(); j++) dlt += vs[j]; vs[id] += vs[id + 1]; // cerr << id << " " << vs.size() << '\n'; vs.erase(vs.begin() + id + 1); int ns = 0; for (int j = 0; j < (signed) lenc.size(); j++) ns += vs[j] * prodc[j]; f[ns] = f[ns] + g[s]; g[s].first += dlt; ns += prodc[id]; f[ns] = f[ns] + g[s]; } } vector<pii> ans (n + 1, pii(inf, 0)); for (int s = 0; s < prodc.back(); s++) { if (f[s].first == inf) continue; vs.clear(); int t = s, len = 0; for (auto l : lenc) len += t % l, t /= l; ans[len] = ans[len] + f[s]; } for (int i = 1; i <= n; i++) { printf("%d %lld\n", ans[i].first, ans[i].second); } return 0; }