結果

問題 No.1332 Range Nearest Query
ユーザー kanra824
提出日時 2021-01-08 22:31:46
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 982 ms / 2,500 ms
コード長 4,537 bytes
コンパイル時間 2,088 ms
コンパイル使用メモリ 205,408 KB
最終ジャッジ日時 2025-01-17 12:49:39
ジャッジサーバーID
(参考情報)
judge4 / judge4
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ファイルパターン 結果
other AC * 48
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ソースコード

diff #
プレゼンテーションモードにする

#include <bits/stdc++.h>
using namespace std;
#define REP(i, n) for(int i=0; i<(n); ++i)
#define RREP(i, n) for(int i=(n);i>=0;--i)
#define FOR(i, a, n) for (int i=(a); i<(n); ++i)
#define RFOR(i, a, b) for(int i=(a);i>=(b);--i)
#define SZ(x) ((int)(x).size())
#define ALL(x) (x).begin(),(x).end()
#define DUMP(x) cerr<<#x<<" = "<<(x)<<endl
#define DEBUG(x) cerr<<#x<<" = "<<(x)<<" (L"<<__LINE__<<")"<<endl;
template<class T>
ostream &operator<<(ostream &os, const vector<T> &v) {
REP(i, SZ(v)) {
if (i) os << " ";
os << v[i];
}
return os;
}
template <class T>
void debug(const vector<T> &v) {
cout << "[";
REP(i, SZ(v)) {
if(i) cout << ", ";
cout << v[i];
}
cout << "]" << endl;
}
template<class T, class U>
ostream &operator<<(ostream &os, const pair<T, U> &p) {
return os << p.first << " " << p.second;
}
template <class T, class U>
void debug(const pair<T, U> &p) {
cout << "(" << p.first << " " << p.second << ")" << endl;
}
template<class T>
bool chmax(T &a, const T &b) {
if (a < b) {
a = b;
return true;
}
return false;
}
template<class T>
bool chmin(T &a, const T &b) {
if (b < a) {
a = b;
return true;
}
return false;
}
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using P = pair<int, int>;
using vi = vector<int>;
using vll = vector<ll>;
using vvi = vector<vi>;
using vvll = vector<vll>;
const ll MOD = 1e9 + 7;
const ll MOD998 = 998244353;
const int INF = INT_MAX;
const ll LINF = LLONG_MAX;
const int inf = INT_MIN;
const ll linf = LLONG_MIN;
const ld eps = 1e-9;
template<int m>
struct mint {
int x;
mint(ll x = 0) : x(((x % m) + m) % m) {}
mint operator-() const { return x ? m - x : 0; }
mint &operator+=(mint r) {
if ((x += r.x) >= m) x -= m;
return *this;
}
mint &operator-=(mint r) {
if ((x -= r.x) < 0) x += m;
return *this;
}
mint &operator*=(mint r) {
x = ((ll) x * r.x) % m;
return *this;
}
mint inv() const { return pow(m - 2); }
mint &operator/=(mint r) { return *this *= r.inv(); }
friend mint operator+(mint l, mint r) { return l += r; }
friend mint operator-(mint l, mint r) { return l -= r; }
friend mint operator*(mint l, mint r) { return l *= r; }
friend mint operator/(mint l, mint r) { return l /= r; }
mint pow(ll n) const {
mint ret = 1, tmp = *this;
while (n) {
if (n & 1) ret *= tmp;
tmp *= tmp, n >>= 1;
}
return ret;
}
friend bool operator==(mint l, mint r) { return l.x == r.x; }
friend bool operator!=(mint l, mint r) { return l.x != r.x; }
friend ostream &operator<<(ostream &os, mint a) {
return os << a.x;
}
friend istream &operator>>(istream &is, mint &a) {
ll x;
is >> x;
a = x;
return is;
}
};
using Int = mint<MOD>;
struct RangeTree {
int n;
vvi node;
RangeTree(vi v) {
n = 1;
int sz = SZ(v);
while(n < sz) {
n *= 2;
}
REP(_, n - sz) {
v.emplace_back(INF);
}
node.resize(2*n+1);
FOR(i, n-1, 2*n-1) {
node[i] = {v[i-n+1]};
}
RREP(i, n-2) {
int l = 2*i+1;
int r = 2*i+2;
vi tmp;
for(auto &e: node[l]) tmp.emplace_back(e);
for(auto &e: node[r]) tmp.emplace_back(e);
sort(ALL(tmp));
node[i] = tmp;
}
}
int query(int l, int r, int a, int b, int k, int x) {
if(b <= l || r <= a) return INF;
if(l <= a && b <= r) {
return mi(k, x);
}
int res1 = query(l, r, a, (a+b)/2, k*2+1, x);
int res2 = query(l, r, (a+b)/2, b, k*2+2, x);
return min(res1, res2);
}
int mi(int idx, int val) {
auto it = lower_bound(ALL(node[idx]), val);
int res = INF;
if(it != node[idx].end()) {
res = abs(val - *it);
}
if(it != node[idx].begin()) {
it--;
chmin(res, abs(val - *it));
}
return res;
}
private:
};
int main() {
cin.tie(0);
ios::sync_with_stdio(false);
cout << fixed << setprecision(10);
int n; cin >> n;
vi v(n);
REP(i, n) cin >> v[i];
RangeTree rt(v);
int q; cin >> q;
while(q--) {
int l, r, x; cin >> l >> r >> x;
l--;
cout << rt.query(l, r, 0, rt.n, 0, x) << endl;
}
return 0;
}
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