結果

問題 No.1115 二つの数列 / Two Sequences
ユーザー null
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
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 5
other AC * 35
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

/*
!
Be accepted!
_ 
。・ω・。)━☆*
      +.
    °+ *´¨)
  .· ´¸.·*´¨) ¸.·*¨)
          (¸.·´ (¸.·'* ☆
*/
#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>
/*/cpp_int
#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)//
#define printdbd(a) printf("%.16lf\n", a)//()
#define prints(s) printf("%s\n", s.c_str())//string
#define all(x) (x).begin(), (x).end()
#define allsum(a, b, c) ((a + b) * c / 2)//,,
#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)); }
/*----------------------------------------------------------------------------------*/
///0-indexed//(, )
template<typename T>
struct segtree {
//
vector<T> seg;
//1/2
int siz;
//
const T unit;
//
using F = function<T(T, T)>;
//
const F f;
//n , a () segtree
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) t
void set(int k, const T& t) {
seg[k + siz] = t;
}
//f
void build() {
for (int i = siz - 1; i >= 0; --i) seg[i] = f(seg[i * 2 + 1], seg[i * 2 + 2]);
}
//i
T operator[](const int i) {
return seg[i + siz];
}
//k a ,使
void update(int k, T a) {
k += siz;
//
seg[k] = a;
while (k > 0) {
k = ((k - 1) >> 1);
seg[k] = f(seg[k * 2 + 1], seg[k * 2 + 2]);
}
}
//[a, b) f
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) [l, r) 調 k check true left, false right f
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';
//
//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;
}
//
if (k >= siz) {
return k - siz;
}
int res;
if (b) {
//調
res = find< C >(start, end, l, ((l + r) >> 1), (k << 1) + 1, check, checknum, b, revf);
//
if (res != -1)return (res);
return find< C >(start, end, ((l + r) >> 1), r, (k << 1) + 2, check, checknum, b, revf);
}
else {
//調
res = find< C >(start, end, ((l + r) >> 1), r, (k << 1) + 2, check, checknum, b, revf);
//
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;
}
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