結果
問題 | No.1115 二つの数列 / Two Sequences |
ユーザー |
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提出日時 | 2020-02-02 23:20:53 |
言語 | C++17(clang) (17.0.6 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 50 ms / 2,000 ms |
コード長 | 6,318 bytes |
コンパイル時間 | 1,655 ms |
コンパイル使用メモリ | 156,812 KB |
実行使用メモリ | 5,632 KB |
最終ジャッジ日時 | 2024-11-30 16:22:06 |
合計ジャッジ時間 | 4,333 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge3 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 5 |
other | AC * 35 |
ソースコード
/*このコード、と~おれ!Be accepted!∧_∧(。・ω・。)つ━☆・*。⊂ ノ ・゜+.しーJ °。+ *´¨).· ´¸.·*´¨) ¸.·*¨)(¸.·´ (¸.·'* ☆*/#include <stdio.h>#include <algorithm>#include <string>#include <cmath>#include <cstring>#include <vector>#include <numeric>#include <iostream>#include <random>#include <map>#include <unordered_map>#include <queue>#include <regex>#include <functional>#include <complex>#include <list>#include <cassert>#include <iomanip>#include <set>/*多倍長整数/cpp_intで宣言#include <boost/multiprecision/cpp_int.hpp>using namespace boost::multiprecision;*///#pragma GCC target ("avx2")//#pragma GCC optimization ("O3")//#pragma GCC optimization ("unroll-loops")#define rep(i, n) for(int i = 0; i < (n); ++i)#define rep1(i, n) for(int i = 1; i <= (n); ++i)#define rep2(i, n) for(int i = 2; i < (n); ++i)#define repr(i, n) for(int i = n; i >= 0; --i)#define reprm(i, n) for(int i = n - 1; i >= 0; --i)#define printynl(a) printf(a ? "yes\n" : "no\n")#define printyn(a) printf(a ? "Yes\n" : "No\n")#define printYN(a) printf(a ? "YES\n" : "NO\n")#define printim(a) printf(a ? "possible\n" : "imposible\n")#define printdb(a) printf("%.50lf\n", a)//少数出力#define printdbd(a) printf("%.16lf\n", a)//少数出力(桁少なめ)#define prints(s) printf("%s\n", s.c_str())//string出力#define all(x) (x).begin(), (x).end()#define allsum(a, b, c) ((a + b) * c / 2)//等差数列の和、初項,末項,項数#define pb push_back#define priq priority_queue#define rpriq priq<int, vector<int>, greater<int>>#define deg_to_rad(deg) (((deg)/360.0)*2.0*PI)#define rad_to_deg(rad) (((rad)/2.0/PI)*360.0)#define Please return#define AC 0#define addf(T) [](T a, T b){return (a + b);}#define minf(T) [](T a, T b){return min(a, b);}#define maxf(T) [](T a, T b){return max(a, b);}using ll = long long;constexpr int INF = 1073741823;constexpr int MINF = -1073741823;constexpr ll LINF = ll(4661686018427387903);constexpr ll MOD = 1000000007;const long double PI = acos(-1.0L);using namespace std;void scans(string& str) {char c;str = "";scanf("%c", &c);if (c == '\n')scanf("%c", &c);while (c != '\n' && c != -1) {str += c;scanf("%c", &c);}}void scanc(char& str) {char c;scanf("%c", &c);if (c == -1)return;while (c == '\n') {scanf("%c", &c);}str = c;}double acot(double x) {return PI / 2 - atan(x);}ll gcd(ll a, ll b) {if (b == 0) return a;return gcd(b, a % b);}ll lcm(ll number1, ll number2) {return number1 / gcd(number1, number2) * number2;}ll LSB(ll n) { return (n & (-n)); }/*-----------------------------------------ここからコード-----------------------------------------*///セグ木/0-indexed/非再帰/(大きさ, 単位元)で初期化template<typename T>struct segtree {//木を配列であらわしたものvector<T> seg;//木の1/2の大きさint siz;//単位元const T unit;//比較関数の型using F = function<T(T, T)>;//マージする関数const F f;//n の大きさ, a (単位元) で segtree を初期化するsegtree(int n, const T a, const F f) : unit(a), f(f) {siz = 1;while (siz < n)siz <<= 1;seg.assign(2 * siz - 1, unit);--siz;}//k (0-indexed) 番目に t を代入void set(int k, const T& t) {seg[k + siz] = t;}//f によって木を構築void build() {for (int i = siz - 1; i >= 0; --i) seg[i] = f(seg[i * 2 + 1], seg[i * 2 + 2]);}//i 番目の要素を返すT operator[](const int i) {return seg[i + siz];}//k 番目の値を a に更新,更新に使う二項演算void update(int k, T a) {k += siz;//必要であればここを変えるseg[k] = a;while (k > 0) {k = ((k - 1) >> 1);seg[k] = f(seg[k * 2 + 1], seg[k * 2 + 2]);}}//[a, b) について f した結果を返すT query(int a, int b) {T l = unit, r = unit;for (a += siz, b += siz; a < b; a >>= 1, b >>= 1) {if (!(a & 1))l = f(l, seg[a++]);if (!(b & 1))r = f(seg[--b], r);}return f(l, r);}//[start, end) について、[l, r) を調べながら k 番目が check を満たすか二分探索 最後が true なら left, false なら right fの逆演算template<typename C>int find(const int start, const int end, int l, int r, int k, const C check, T& checknum, const bool b, const function<T(T, T)> revf) {//cerr << checknum << '\n';//範囲外またはそこがすでに満たさないとき//cerr << k << ',' << checknum << '\n';if (start <= l && r <= end && !check(seg[k], checknum)) {checknum = revf(checknum, seg[k]);return -1;}if ((r <= start || l >= end)) {return -1;}//既に葉if (k >= siz) {return k - siz;}int res;if (b) {//左側を調べるres = find< C >(start, end, l, ((l + r) >> 1), (k << 1) + 1, check, checknum, b, revf);//左側が適してたらそれが答えif (res != -1)return (res);return find< C >(start, end, ((l + r) >> 1), r, (k << 1) + 2, check, checknum, b, revf);}else {//右側を調べるres = find< C >(start, end, ((l + r) >> 1), r, (k << 1) + 2, check, checknum, b, revf);//右側が適してたらそれが答えif (res != -1)return (res);return find< C >(start, end, l, ((l + r) >> 1), (k << 1) + 1, check, checknum, b, revf);}}template<typename C>int find_left(int start, int end, const C check, T& checknum, function<T(T, T)> revf) {return find< C >(start, end, 0, siz + 1, 0, check, checknum, true, revf);}template<typename C>int find_right(int start, int end, const C check, T& checknum, function<T(T, T)> revf) {return find< C >(start, end, 0, siz + 1, 0, check, checknum, false, revf);}};int main() {int n;scanf("%d", &n);segtree<int> tree(n + 1, 0, addf(int));vector<int> a(n);vector<pair<int, int>> b(n);rep(i, n)scanf("%d", &a[i]);rep(i, n) {scanf("%d", &b[i].first);b[i].second = i + 1;}sort(all(b));ll ans = 0;rep(i, n) {a[i] = b[a[i] - 1].second;ans += tree.query(a[i], n + 1);tree.update(a[i], tree[a[i]] + 1);}printf("%lld\n", ans);Please AC;}