結果

問題 No.3101 Range Eratosthenes Query
ユーザー hato336
提出日時 2025-04-12 08:48:29
言語 C++23
(gcc 13.3.0 + boost 1.87.0)
結果
TLE  
実行時間 -
コード長 10,132 bytes
コンパイル時間 4,205 ms
コンパイル使用メモリ 297,168 KB
実行使用メモリ 27,056 KB
最終ジャッジ日時 2025-04-12 08:48:43
合計ジャッジ時間 13,390 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
（要ログイン）
ファイルパターン 結果
sample TLE * 1 -- * 1
other -- * 24
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 # 

#include <bits/stdc++.h>

using namespace std;

#define all(...) std::begin(__VA_ARGS__), std::end(__VA_ARGS__)
#define rall(...) std::rbegin(__VA_ARGS__), std::rend(__VA_ARGS__)
#define OVERLOAD_REP(_1, _2, _3, _4, name, ...) name
#define REP1(n) for(ll i=0;i<n;i++)
#define REP2(i, n) for (ll i=0;i<n;i++)
#define REP3(i, a, n) for (ll i=a;i<n;i++)
#define REP4(i, a, b, n) for(ll i=a;i<n;i+=b)
#define rep(...) OVERLOAD_REP(__VA_ARGS__, REP4, REP3, REP2, REP1)(__VA_ARGS__)
#define OVERLOAD_RREP(_1, _2, _3, _4, name, ...) name
#define RREP1(n) for(ll i=n-1;i>=0;i--)
#define RREP2(i, n) for(ll i=n-1;i>=0;i--)
#define RREP3(i, a, n) for(ll i=n-1;i>=a;i--)
#define RREP4(i, a, b, n) for(ll i=n-1;i>=a;i-=b)
#define rrep(...) OVERLOAD_RREP(__VA_ARGS__, RREP4, RREP3, RREP2, RREP1)(__VA_ARGS__)
#define foa(a,v)  (auto& a : (v))
#define uniq(a) sort(all(a));a.erase(unique(all(a)),end(a))
#define len(n) (long long)(n).size()
#define pb push_back
using ll = long long;
using ld = long double;
using ull = unsigned long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vvvi = vector<vvi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vvvll = vector<vvll>;
using vs = vector<string>;
using vvs = vector<vs>;
using vvvs = vector<vvs>;
using vld = vector<ld>;
using vvld = vector<vld>;
using vvvld = vector<vvld>;
using vc = vector<char>;
using vvc = vector<vc>;
using vvvc = vector<vvc>;
using pll = pair<ll,ll>;
using vpll = vector<pll>;
template<class... T>
constexpr auto min(T... a){
    return min(initializer_list<common_type_t<T...>>{a...});
}

template<class... T>
void input(T&... a){
    (cin >> ... >> a);
}
ll POW(ll a,ll b){
    ll ans = 1;
    while (b){
        if (b & 1){
            ans *= a;
        }
        a *= a;
        b /= 2;
    }
    return ans;
}
ll MODPOW(ll a,ll b,ll c){
    ll ans = 1;
    while (b){
        if (b & 1){
            ans *= a;
            ans %= c;
        }
        a *= a;
        a %= c;
        b /= 2;
    }
    return ans;
}

#define INT(...) int __VA_ARGS__; input(__VA_ARGS__)
#define LL(...) ll __VA_ARGS__; input(__VA_ARGS__)
#define ULL(...) ull __VA_ARGS__; input(__VA_ARGS__)
#define LD(...) ld __VA_ARGS__; input(__VA_ARGS__)
#define STR(...) string __VA_ARGS__; input(__VA_ARGS__)
#define CHA(...) char __VA_ARGS__; input(__VA_ARGS__)
#define VLL(name,length) vll name(length);rep(i,length){cin >> name[i];}
#define VVLL(name,h,w) vvll name(h,vll(w));rep(i,h)rep(j,w){cin >> name[i][j];}
#define VVVLL(name,a,b,c) vvvll name(a,vvll(b,vll(c)));rep(i,a)rep(j,b)rep(k,c){cin >> name[i][j][k];}
#define VI(name,length) vi name(length);rep(i,length){cin >> name[i];}
#define VVI(name,h,w) vvi name(h,vi(w));rep(i,h)rep(j,w){cin >> name[i][j];}
#define VVVI(name,a,b,c) vvvi name(a,vvll(b,vi(c)));rep(i,a)rep(j,b)rep(k,c){cin >> name[i][j][k];}
#define VLD(name,length) vld name(length);rep(i,length){cin >> name[i];}
#define VVLD(name,h,w) vvld name(h,vld(w));rep(i,h)rep(j,w){cin >> name[i][j];}
#define VVVLD(name,a,b,c) vvvld name(a,vvld(b,vld(c)));rep(i,a)rep(j,b)rep(k,c){cin >> name[i][j][k];}
#define VC(name,length) vc name(length);rep(i,length){cin >> name[i];}
#define VVC(name,h,w) vvc name(h,vc(w));rep(i,h)rep(j,w){cin >> name[i][j];}
#define VVVC(name,a,b,c) vvvc name(a,vvc(b,vc(c)));rep(i,a)rep(j,b)rep(k,c){cin >> name[i][j][k];}
#define VS(name,length) vs name(length);rep(i,length){cin >> name[i];}
#define VVS(name,h,w) vvs name(h,vs(w));rep(i,h)rep(j,w){cin >> name[i][j];}
#define VVVS(name,a,b,c) vvvs name(a,vvs(b,vs(c)));rep(i,a)rep(j,b)rep(k,c){cin >> name[i][j][k];}
#define PLL(name) pll name;cin>>name.first>>name.second;
#define VPLL(name,length) vpll name(length);rep(i,length){cin>>name[i].first>>name[i].second;}

void print(){cout << "\n";}
template<class T, class... Ts>
void print(const T& a, const Ts&... b){cout << a;(cout << ... << (cout << ' ', b));cout << '\n';}
void print(vll x){rep(i,len(x)){cout << x[i];if(i!=len(x)-1){cout << " ";}else{cout << '\n';}}}
void print(vvll x){rep(i,len(x))rep(j,len(x[i])){cout << x[i][j];if(j!=len(x[i])-1){cout << " ";}else{cout << '\n';}}}
void print(vi x){rep(i,len(x)){cout << x[i];if(i!=len(x)-1){cout << " ";}else{cout << '\n';}}}
void print(vvi x){rep(i,len(x))rep(j,len(x[i])){cout << x[i][j];if(j!=len(x[i])-1){cout << " ";}else{cout << '\n';}}}
void print(vvvi x){rep(i,len(x))rep(j,len(x[i]))rep(k,len(x[i][j])){cout << x[i][j][k];if(k!=len(x[i][j])-1){cout << " ";}else if(j!=len(x[i])-1){cout << " | ";}else{cout << '\n';}}}
void print(vld x){rep(i,len(x)){cout << x[i];if(i!=len(x)-1){cout << " ";}else{cout << '\n';}}}
void print(vvld x){rep(i,len(x))rep(j,len(x[i])){cout << x[i][j];if(j!=len(x[i])-1){cout << " ";}else{cout << '\n';}}}
void print(vvvld x){rep(i,len(x))rep(j,len(x[i]))rep(k,len(x[i][j])){cout << x[i][j][k];if(k!=len(x[i][j])-1){cout << " ";}else if(j!=len(x[i])-1){cout << " | ";}else{cout << '\n';}}}
void print(vc x){rep(i,len(x)){cout << x[i];if(i!=len(x)-1){cout << " ";}else{cout << '\n';}}}
void print(vvc x){rep(i,len(x))rep(j,len(x[i])){cout << x[i][j];if(j!=len(x[i])-1){cout << " ";}else{cout << '\n';}}}
void print(vvvc x){rep(i,len(x))rep(j,len(x[i]))rep(k,len(x[i][j])){cout << x[i][j][k];if(k!=len(x[i][j])-1){cout << " ";}else if(j!=len(x[i])-1){cout << " | ";}else{cout << '\n';}}}
void print(vs x){rep(i,len(x)){cout << x[i];if(i!=len(x)-1){cout << " ";}else{cout << '\n';}}}
void print(vvs x){rep(i,len(x))rep(j,len(x[i])){cout << x[i][j];if(j!=len(x[i])-1){cout << " ";}else{cout << '\n';}}}
void print(vvvs x){rep(i,len(x))rep(j,len(x[i]))rep(k,len(x[i][j])){cout << x[i][j][k];if(k!=len(x[i][j])-1){cout << " ";}else if(j!=len(x[i])-1){cout << " | ";}else{cout << '\n';}}}
void print(pll x){cout << x.first << x.second << '\n';}
void print(vpll x){rep(i,len(x)){cout << x[i].first << x[i].second << '\n';}}

#line 2 "misc/mo.hpp"

struct Mo {
  int width;
  vector<int> left, right, order;

  Mo(int N, int Q) : order(Q) {
    width = max<int>(1, 1.0 * N / max<double>(1.0, sqrt(Q * 2.0 / 3.0)));
    iota(begin(order), end(order), 0);
  }

  void insert(int l, int r) { /* [l, r) */
    left.emplace_back(l);
    right.emplace_back(r);
  }

  template <typename AL, typename AR, typename DL, typename DR, typename REM>
  void run(const AL &add_left, const AR &add_right, const DL &delete_left,
           const DR &delete_right, const REM &rem) {
    assert(left.size() == order.size());
    sort(begin(order), end(order), [&](int a, int b) {
      int ablock = left[a] / width, bblock = left[b] / width;
      if (ablock != bblock) return ablock < bblock;
      if (ablock & 1) return right[a] < right[b];
      return right[a] > right[b];
    });
    int nl = 0, nr = 0;
    for (auto idx : order) {
      while (nl > left[idx]) add_left(--nl);
      while (nr < right[idx]) add_right(nr++);
      while (nl < left[idx]) delete_left(nl++);
      while (nr > right[idx]) delete_right(--nr);
      rem(idx);
    }
  }
};

/**
 * @brief Mo's algorithm
 * @docs docs/misc/mo.md
 */




// CUT begin
// Count elements in $[A_\mathrm{begin}, ..., A_{\mathrm{end}-1}]$ which satisfy $A_i < \mathrm{query}$
// Complexity: $O(N \log N)$ for initialization, $O(\log^2 N)$ for each query
// Verified: https://codeforces.com/contest/1288/submission/68865506
template <typename T> struct merge_sort_tree {
    int N;
    std::vector<std::vector<T>> x;
    merge_sort_tree() = default;
    merge_sort_tree(const std::vector<T> &vec) : N(vec.size()) {
        x.resize(N * 2);
        for (int i = 0; i < N; i++) x[N + i] = {vec[i]};
        for (int i = N - 1; i; i--) {
            std::merge(x[i * 2].begin(), x[i * 2].end(), x[i * 2 + 1].begin(), x[i * 2 + 1].end(),
                       std::back_inserter(x[i]));
        }
    }
    int cntLess(int l, int r, T query) const {
        l += N, r += N;
        int ret = 0;
        while (l < r) {
            if (l & 1)
                ret += std::lower_bound(x[l].begin(), x[l].end(), query) - x[l].begin(), l++;
            if (r & 1)
                r--, ret += std::lower_bound(x[r].begin(), x[r].end(), query) - x[r].begin();
            l >>= 1, r >>= 1;
        }
        return ret;
    }
    int cntLesseq(int l, int r, T query) const {
        l += N, r += N;
        int ret = 0;
        while (l < r) {
            if (l & 1)
                ret += std::upper_bound(x[l].begin(), x[l].end(), query) - x[l].begin(), l++;
            if (r & 1)
                r--, ret += std::upper_bound(x[r].begin(), x[r].end(), query) - x[r].begin();
            l >>= 1, r >>= 1;
        }
        return ret;
    }
    int cntMore(int begin, int end, T query) const {
        int tot = std::max(0, std::min(end, N) - std::max(begin, 0));
        return tot - cntLesseq(begin, end, query);
    }
    int cntMoreeq(int begin, int end, T query) const {
        int tot = std::max(0, std::min(end, N) - std::max(begin, 0));
        return tot - cntLess(begin, end, query);
    }

    template <class OStream> friend OStream &operator<<(OStream &os, const merge_sort_tree &clt) {
        os << '[';
        for (int i = 0; i < clt.N; i++) os << clt.x[clt.N + i][0] << ',';
        return os << ']';
    }
};


// N の約数をすべて求める関数
vector<long long> calc_divisors(long long N) {
    // 答えを表す集合
    vector<long long> res;

    // 各整数 i が N の約数かどうかを調べる
    for (long long i = 1; i * i <= N; ++i) {
        // i が N の約数でない場合はスキップ
        if (N % i != 0) continue;

        // i は約数である
        res.push_back(i);

        // N ÷ i も約数である (重複に注意)
        if (N / i != i) res.push_back(N / i);
    }

    // 約数を小さい順に並び替えて出力
    sort(res.begin(), res.end());
    return res;
}



int main(){
	ios::sync_with_stdio(false);
	std::cin.tie(nullptr);
    LL(q);
    ll m = 1000000+10;
    vll a(m,0);
    a[1] = 0;
    rep(i,2,m){
        auto x = calc_divisors(i);
        a[i] = x[len(x)-2];
    }
    merge_sort_tree<ll> st(a);
    rep(i,q){
        LL(l,r);
        ll x = st.cntMoreeq(l,r+1,l);
        print(r-l+1-x);
    }
    
}
0