結果

問題 No.3051 Make All Divisible
ユーザー shobonvip
提出日時 2025-03-07 23:27:26
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
TLE  
実行時間 -
コード長 3,929 bytes
コンパイル時間 5,140 ms
コンパイル使用メモリ 271,852 KB
実行使用メモリ 118,272 KB
最終ジャッジ日時 2025-03-07 23:27:37
合計ジャッジ時間 10,585 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge2
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
other AC * 23 TLE * 1 -- * 7
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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

/**
    author:  shobonvip
    created: 2025.03.07 22:05:11
**/
#include<bits/stdc++.h>
using namespace std;
//* ATCODER
#include<atcoder/all>
using namespace atcoder;
typedef modint998244353 mint;
//*/
/* BOOST MULTIPRECISION
#include<boost/multiprecision/cpp_int.hpp>
using namespace boost::multiprecision;
//*/
typedef long long ll;
#define rep(i, s, n) for (int i = (int)(s); i < (int)(n); i++)
#define rrep(i, s, n) for (int i = (int)(n)-1; i >= (int)(s); i--)
#define all(v) v.begin(), v.end()
template <typename T> bool chmin(T &a, const T &b) {
    if (a <= b) return false;
    a = b;
    return true;
}
template <typename T> bool chmax(T &a, const T &b) {
    if (a >= b) return false;
    a = b;
    return true;
}
template <typename T> T max(vector<T> &a){
    assert(!a.empty());
    T ret = a[0];
    for (int i=0; i<(int)a.size(); i++) chmax(ret, a[i]);
    return ret;
}
template <typename T> T min(vector<T> &a){
    assert(!a.empty());
    T ret = a[0];
    for (int i=0; i<(int)a.size(); i++) chmin(ret, a[i]);
    return ret;
}
template <typename T> T sum(vector<T> &a){
    T ret = 0;
    for (int i=0; i<(int)a.size(); i++) ret += a[i];
    return ret;
}
ll op(ll a,ll b) {
    return a + b;
}
ll e(){
    return 0;
}
void solve() {
    ll k, n;
    cin >> n >> k;
    vector<ll> a(n);
    vector<ll> b(n);
    vector<ll> c(n);
    ll s = 0;
    rep(i,0,n) {
        cin >> a[i];
        b[i] = a[i] - a[i] / k * k;
        c[i] = a[i] / k;
        s += b[i];
    }
    vector<ll> d = c;
    sort(all(d));
    if (s % k != 0) {
        cout << -1 << '\n';
        return;
    }
    if (s == 0) {
        cout << 0 << '\n';
        return;
    }
    auto check = [&](ll i) {
        ll p = i + s/k;
        ll freedom = 0;
        ll cost = 0;
        vector<ll> amari(k);
        rep(i,0,n) {
            if (b[i] > p) {
                cost += p;
            }else {
                freedom += min(c[i], (p - b[i]) / k);
                if (a[i] > p) {
                    amari[(p - b[i]) % k] += 1;
                }
                cost += b[i];
            }
        }
        ll left = i;
        {
            ll targ = min(freedom, left);
            freedom -= targ;
            left -= targ;
            cost += targ * k;
        }
        rrep(i,0,k) { 
            ll targ = min(amari[i], left);
            left -= targ;
            amari[i] -= targ;
            cost += targ * i;
        }
        if (p*k<=cost) return true;
        return false;
    };
    for(ll i=0;i<=min(sum(c),k);i++) {
        if (check(i)) {
            cout << i + s/k << '\n';
            return;
        }
    }
    ll bsum = sum(b);
    vector<segtree<ll,op,e>> seg1(k, segtree<ll,op,e>(k+1));
    vector<segtree<ll,op,e>> seg2(k, segtree<ll,op,e>(k+1));
    vector<bool> seen(k);
    ll tmp;
    auto f = [&](ll x) -> bool {
        return x <= tmp;
    };
    
    vector ri(k, vector<pair<ll,ll>>(n));
    vector<ll> piv(k);
    for (ll i=k; i<=min(sum(c),n*k+10); i++) {
        ll p = i+s/k;
        int var=p%k;
        if (!seen[var]){
            rep(x,0,n){
                ri[var][x] = pair(c[x]-(p-b[x])/k, (p-b[x])%k);
                seg1[var].set(k, seg1[var].get(k) + min(c[x],(p-b[x])/k));
                seg2[var].set(k, seg2[var].get(k) + min(c[x],(p-b[x])/k) * k);
                // not c[x] <= (p-b[x])/k
                // <=> c[x] - (p-b[x])/k > 0
                if (c[x]*k+b[x] > p) {
                    ll am = (p-b[x])%k;
                    seg1[var].set(am, seg1[var].get(am) + 1);
                    seg2[var].set(am, seg2[var].get(am) + am);
                }
            }
            sort(all(ri[var]));
            while(piv[var]<n && ri[var][piv[var]].first <= i/k-1) piv[var]++;
            seen[var] = 1;
        }else{
            seg1[var].set(k, seg1[var].get(k) + (n - piv[var]));
            seg2[var].set(k, seg2[var].get(k) + (n - piv[var]) * k);
            // now ... c[i] - (p-b[i])/k >= 0
            while(piv[var]<n && ri[var][piv[var]].first <= i/k-1) {
                ll am = ri[var][piv[var]].second;
                seg1[var].set(am, seg1[var].get(am) - 1);
                seg2[var].set(am, seg2[var].get(am) - am);
                piv[var]++; 
            }
            // now ... c[i] - (p-b[i])/k >= 1
        }
        tmp = i;
        ll cost = bsum;
        ll v = seg1[var].min_left(k+1, f);
        tmp -= seg1[var].prod(v, k+1);
        cost += seg2[var].prod(v, k+1);
        assert(tmp >= 0);
        if (v-1 >= 0){
            cost += (v-1) * tmp;
        }
        if (p*k<=cost) {
            cout << p << '\n';
            return;
        }
    } 
    cout << -1 << '\n';
    return;
    
}
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    
    int t; cin >> t;
    while(t--) solve();
}
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