結果
問題 | No.2501 Maximum Inversion Number |
ユーザー |
|
提出日時 | 2023-10-14 00:10:21 |
言語 | C++23 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 390 ms / 2,000 ms |
コード長 | 4,221 bytes |
コンパイル時間 | 5,719 ms |
コンパイル使用メモリ | 315,080 KB |
実行使用メモリ | 7,936 KB |
最終ジャッジ日時 | 2024-09-15 20:03:09 |
合計ジャッジ時間 | 7,874 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge2 |
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ファイルパターン | 結果 |
---|---|
other | AC * 17 |
ソースコード
#include "bits/stdc++.h"#include <unordered_map>#include <unordered_set>#include <stdio.h>#include <math.h>#include <cassert>#include "atcoder/all"#include <float.h>//#include <boost/multiprecision/cpp_int.hpp>//using namespace boost::multiprecision;// std::fixed << std::setprecision(10) <<using namespace std;using namespace atcoder;using ll = long long;using ull = unsigned long long;using mint = modint998244353;using mint2 = modint1000000007;const double eps = 1e-9;#define REP(i, n) for (ll i = 0; i < ll(n); ++i)#define REPI(i, init, n) for (ll i = init; i < ll(n); ++i)#define REPD(i, init) for (ll i = init; i >=0; --i)#define REPDI(i, init, n) for (ll i = init; i >=n; --i)#define REPE(i, c) for (const auto& i : c)#define TCASE() ll _ttt; cin >> _ttt; while(_ttt--)using vl = vector<ll>;#define VL(a, n) vector<ll> a; a.assign(n, 0);#define VLI(a, n) vector<ll> a; a.assign(n, 0); for(auto& x : a) cin >> x;#define VSI(a, n) vector<string> a; a.assign(n, ""); for(auto& x : a) cin >> x;using vvl = vector<vector<ll>>;#define VVLI(a, n, m, init) vector<vector<ll>> a(n); for(auto& x : a) x.assign(m, init);using vd = vector<double>;using pl = pair<ll, ll>;struct uv { ll u; ll v; ll c; };#define VUVI(a, n) vector<uv> a; a.assign(n, {0, 0, 1}); for(auto& x : a) {cin >> x.u >> x.v; x.u--; x.v--;}#define VUVCI(a, n) vector<uv> a; a.assign(n, {0, 0, 0}); for(auto& x : a) {cin >> x.u >> x.v >> x.c; x.u--; x.v--;}vvl to_edge(const ll n, const vector<uv>& v) { vvl ret(n); for (auto& x : v) ret[x.u].push_back(x.v); return ret; }vvl to_edge_d(const ll n, const vector<uv>& v) { vvl ret(n); for (auto& x : v) { ret[x.u].push_back(x.v); ret[x.v].push_back(x.u); } return ret; }template <class T = long long> using pql = priority_queue<T>;template <class T = long long> using pqg = priority_queue<T, vector<T>, greater<T>>;using vm = vector<mint>;using vvm = vector<vm>;void yn(bool f) { std::cout << (f ? "Yes" : "No") << endl; };template<class T> void ov(const T& v) { for (auto it = v.begin(); it != v.end(); it++) { if (it != v.begin()) cout << " "; cout << *it; } };template<> void ov(const vm& v) { for (auto it = v.begin(); it != v.end(); it++) { if (it != v.begin()) cout << " "; cout << it->val(); } };const ll llhuge = 1LL << 60; //十分でかいが多少足しても溢れない数値auto gmax(const auto& a, const auto& b) { return a > b ? a : b; };auto gmin(const auto& a, const auto& b) { return a < b ? a : b; };template<class T, class U> bool chmin(T& a, const U& b) { const T olda = a; a = gmin(a, b); return olda != a; }template<class T, class U> bool chmax(T& a, const U& b) { const T olda = a; a = gmax(a, b); return olda != a; }template<class T> void mysort(T& v) { std::sort(begin(v), end(v)); };template<class T, class U> void mysort(T& v, U pr) { std::sort(begin(v), end(v), pr); };template<class T> void myrev(T& v) { std::reverse(begin(v), end(v)); };//二分探索でf(t)を満たす整数tを返す//ng < ok ならtは最小値 ng > ok ならtは最大値template <class F>ll bs(ll ng, ll ok, F f) {while (abs(ok - ng) > 1) {ll mid = (ok + ng) / 2;if (f(mid)) ok = mid;else ng = mid;}return ok;}int main() {cin.tie(nullptr);ios::sync_with_stdio(false);TCASE() {ll n, m;cin >> n >> m;VLI(l, n);VLI(r, n);ll lsum = 0;ll rsum = 0;REP(i, n) {lsum += l[i];rsum += r[i];}if (lsum > m || rsum < m) {cout << -1 << endl;continue;}//全ての要素が可能なら満たす数字を調べるll tv = bs(m + 1, 0, [&](ll t) {ll sum = 0;REP(i, n) {ll add = t;if (r[i] < t) add = r[i];if (l[i] > t) add = l[i];sum += add;}return sum <= m;});//全ての要素を弄ってしまう。vl tl = l;REP(i, n) {if (r[i] < tv) tl[i] = r[i];else chmax(tl[i], tv);m -= tl[i];}//mの残りは大きいほうから1ずつ足せれば足すREPD(i, n - 1) {if (m <= 0) break;if (tl[i] == tv && tl[i] < r[i]) {tl[i]++;m--;}}//転倒数を数えるll ans = 0;ll revt = 0;REP(i, n) {ans += revt * tl[i];revt += tl[i];}cout << ans << endl;}}