結果

問題 No.3078 Difference Sum Query
ユーザー 👑 binap
提出日時 2025-03-29 10:09:07
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 1,465 ms / 2,000 ms
コード長 3,176 bytes
コンパイル時間 5,602 ms
コンパイル使用メモリ 289,908 KB
実行使用メモリ 7,328 KB
最終ジャッジ日時 2025-03-29 10:09:42
合計ジャッジ時間 32,447 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge4
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 * 2
other AC * 26
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

#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
#include<atcoder/all>
#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;
using namespace atcoder;
typedef long long ll;
typedef vector<int> vi;
typedef vector<long long> vl;
typedef vector<vector<int>> vvi;
typedef vector<vector<long long>> vvl;
typedef long double ld;
typedef pair<int, int> P;
template <int m> ostream& operator<<(ostream& os, const static_modint<m>& a) {os << a.val(); return os;}
template <int m> ostream& operator<<(ostream& os, const dynamic_modint<m>& a) {os << a.val(); return os;}
template <int m> istream& operator>>(istream& is, static_modint<m>& a) {long long x; is >> x; a = x; return is;}
template <int m> istream& operator>>(istream& is, dynamic_modint<m>& a) {long long x; is >> x; a = x; return is;}
template<typename T> istream& operator>>(istream& is, vector<T>& v){int n = v.size(); assert(n > 0); rep(i, n) is >> v[i]; return is;}
template<typename U, typename T> ostream& operator<<(ostream& os, const pair<U, T>& p){os << p.first << ' ' << p.second; return os;}
template<typename T> ostream& operator<<(ostream& os, const vector<T>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : " "); return 
    os;}
template<typename T> ostream& operator<<(ostream& os, const vector<vector<T>>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : ""); 
    return os;}
template<typename T> ostream& operator<<(ostream& os, const set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename T> ostream& operator<<(ostream& os, const unordered_set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename S, auto op, auto e> ostream& operator<<(ostream& os, const atcoder::segtree<S, op, e>& seg){int n = seg.max_right(0, [](S){return 
    true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;}
template<typename S, auto op, auto e, typename F, auto mapping, auto composition, auto id> ostream& operator<<(ostream& os, const atcoder
    ::lazy_segtree<S, op, e, F, mapping, composition, id>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 
    1 ? "\n" : " "); return os;}
template<typename T> void chmin(T& a, T b){a = min(a, b);}
template<typename T> void chmax(T& a, T b){a = max(a, b);}
int main(){
    int n, q;
    cin >> n >> q;
    vector<long long> a(n);
    cin >> a;
    
    vector<long long> b = a;
    sort(b.begin(), b.end());
    vector<long long> b_sum(n + 1);
    rep(i, n) b_sum[i + 1] = b_sum[i] + b[i];
    
    auto calc = [&](long long x){
        int idx = lower_bound(b.begin(), b.end(), x) - b.begin();
        long long res = 0;
        res += b_sum[n] - b_sum[idx];
        res -= (n - idx) * x;
        res -= b_sum[idx] - b_sum[0];
        res += idx * x;
        return res;
    };
    
    rep(i, q){
        int l, r;
        long long x;
        cin >> l >> r >> x;
        l--; r--;
        
        long long ans = 0;
        if(r - l <= n / 2){
            for(int j = l; j <= r; j++){
                ans += abs(a[j] - x);
            }
        }else{
            ans += calc(x);
            for(int j = 0; j <= l - 1; j++){
                ans -= abs(a[j] - x);
            }
            for(int j = r + 1; j <= n - 1; j++){
                ans -= abs(a[j] - x);
            }
        }
        cout << ans << "\n";
    }
    
    return 0;
}
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