結果

問題 No.1022 Power Equation
ユーザー hitonanodehitonanode
提出日時 2020-04-10 21:37:16
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 78 ms / 2,000 ms
コード長 6,046 bytes
コンパイル時間 1,921 ms
コンパイル使用メモリ 177,408 KB
実行使用メモリ 5,376 KB
最終ジャッジ日時 2024-09-15 19:31:38
合計ジャッジ時間 2,782 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge6
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 * 1
other AC * 8
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;
using lint = long long int;
using pint = pair<int, int>;
using plint = pair<lint, lint>;
struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
#define ALL(x) (x).begin(), (x).end()
#define FOR(i, begin, end) for(int i=(begin),i##_end_=(end);i<i##_end_;i++)
#define IFOR(i, begin, end) for(int i=(end)-1,i##_begin_=(begin);i>=i##_begin_;i--)
#define REP(i, n) FOR(i,0,n)
#define IREP(i, n) IFOR(i,0,n)
template<typename T> void ndarray(vector<T> &vec, int len) { vec.resize(len); }
template<typename T, typename... Args> void ndarray(vector<T> &vec, int len, Args... args) { vec.resize(len); for (auto &v : vec) ndarray(v, args...); }
template<typename T> bool chmax(T &m, const T q) { if (m < q) {m = q; return true;} else return false; }
template<typename T> bool chmin(T &m, const T q) { if (m > q) {m = q; return true;} else return false; }
template<typename T1, typename T2> pair<T1, T2> operator+(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first + r.first, l.second + r.second); }
template<typename T1, typename T2> pair<T1, T2> operator-(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first - r.first, l.second - r.second); }
template<typename T> istream &operator>>(istream &is, vector<T> &vec){ for (auto &v : vec) is >> v; return is; }
template<typename T> ostream &operator<<(ostream &os, const vector<T> &vec){ os << "["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template<typename T> ostream &operator<<(ostream &os, const deque<T> &vec){ os << "deq["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template<typename T> ostream &operator<<(ostream &os, const set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const unordered_set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const unordered_multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &pa){ os << "(" << pa.first << "," << pa.second << ")"; return os; }
template<typename TK, typename TV> ostream &operator<<(ostream &os, const map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }
template<typename TK, typename TV> ostream &operator<<(ostream &os, const unordered_map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }
#define dbg(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ") " << __FILE__ << endl;
/*
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/tag_and_trait.hpp>
using namespace __gnu_pbds; // find_by_order(), order_of_key()
template<typename TK> using pbds_set = tree<TK, null_type, less<TK>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename TK, typename TV> using pbds_map = tree<TK, TV, less<TK>, rb_tree_tag, tree_order_statistics_node_update>;
*/

// Sieve of Eratosthenes
// (*this)[i] = (divisor of i, greater than 1)
// Example: [0, 1, 2, 3, 2, 5, 3, 7, 2, 3, 2, 11, ...]
// Complexity: Space O(MAXN), Time (construction) O(MAXNloglogMAXN)
struct SieveOfEratosthenes : std::vector<int>
{
    std::vector<int> primes;
    SieveOfEratosthenes(int MAXN) : std::vector<int>(MAXN + 1) {
        std::iota(begin(), end(), 0);
        for (int i = 2; i <= MAXN; i++) {
            if ((*this)[i] == i) {
                primes.push_back(i);
                for (int j = i; j <= MAXN; j += i) (*this)[j] = i;
            }
        }
    }
    using T = long long int;
    // Prime factorization for x <= MAXN^2
    // Complexity: O(log x)          (x <= MAXN)
    //             O(MAXN / logMAXN) (MAXN < x <= MAXN^2)
    std::map<T, int> Factorize(T x) {
        assert(x <= 1LL * (int(size()) - 1) * (int(size()) - 1));
        std::map<T, int> ret;
        if (x < int(size())) {
            while (x > 1) {
                ret[(*this)[x]]++;
                x /= (*this)[x];
            }
        }
        else {
            for (auto p : primes) {
                while (!(x % p)) x /= p, ret[p]++;
                if (x == 1) break;
            }
            if (x > 1) ret[x]++;
        }
        return ret;
    }
    std::vector<T> Divisors(T x) {
        std::vector<T> ret{1};
        for (auto p : Factorize(x)) {
            int n = ret.size();
            for (int i = 0; i < n; i++) {
                for (T a = 1, d = 1; d <= p.second; d++) {
                    a *= p.first;
                    ret.push_back(ret[i] * a);
                }
            }
        }
        return ret; // Not sorted
    }
    // Moebius function Table
    // return: [0=>0, 1=>1, 2=>-1, 3=>-1, 4=>0, 5=>-1, 6=>1, 7=>-1, 8=>0, ...]
    std::vector<int> GenerateMoebiusFunctionTable() {
        std::vector<int> ret(size());
        for (int i = 1; i < int(size()); i++) {
            if (i == 1) ret[i] = 1;
            else if ((i / (*this)[i]) % (*this)[i] == 0) ret[i] = 0;
            else ret[i] = -ret[i / (*this)[i]];
        }
        return ret;
    }
};
SieveOfEratosthenes sieve(40101);

int main()
{
    int T;
    cin >> T;
    while (T--)
    {
        lint N;
        cin >> N;
        lint ret = (N - 1) * N;  // a == c (a > 1)
        ret += N * N;  // a == c == 1
        for (lint a = 2; a * a <= N; a++)
        {
            lint c = a;
            int deg = 1;
            while (c * a <= N) deg++, c *= a;
            FOR(d1, 1, deg + 1) FOR(d2, 1, deg + 1) if (__gcd(d1, d2) == 1 and d1 != d2)
            {
                ret += N / max(d1, d2);
            }
        }
        cout << ret << "\n";
    }
}
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