結果

問題 No.2187 三立法和 mod 333
ユーザー dyktr_06dyktr_06
提出日時 2023-01-13 22:21:43
言語 C++17
(gcc 12.3.0 + boost 1.83.0)
結果
TLE  
実行時間 -
コード長 5,242 bytes
コンパイル時間 6,148 ms
コンパイル使用メモリ 302,440 KB
実行使用メモリ 6,944 KB
最終ジャッジ日時 2024-06-06 23:08:14
合計ジャッジ時間 9,682 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge2
このコードへのチャレンジ
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 TLE -
testcase_01 AC 71 ms
5,376 KB
testcase_02 AC 48 ms
5,376 KB
testcase_03 AC 35 ms
5,376 KB
testcase_04 AC 34 ms
5,376 KB
testcase_05 AC 48 ms
5,376 KB
testcase_06 AC 72 ms
5,376 KB
testcase_07 TLE -
testcase_08 TLE -
testcase_09 TLE -
testcase_10 AC 73 ms
5,376 KB
testcase_11 AC 79 ms
5,376 KB
testcase_12 AC 99 ms
5,376 KB
testcase_13 AC 79 ms
5,376 KB
testcase_14 AC 50 ms
5,376 KB
testcase_15 TLE -
testcase_16 AC 70 ms
5,376 KB
testcase_17 AC 77 ms
5,376 KB
testcase_18 AC 34 ms
5,376 KB
testcase_19 TLE -
testcase_20 TLE -
testcase_21 TLE -
testcase_22 TLE -
testcase_23 AC 36 ms
5,376 KB
testcase_24 TLE -
testcase_25 AC 36 ms
5,376 KB
testcase_26 AC 36 ms
5,376 KB
testcase_27 TLE -
testcase_28 AC 49 ms
5,376 KB
testcase_29 TLE -
testcase_30 AC 47 ms
5,376 KB
testcase_31 AC 47 ms
5,376 KB
testcase_32 TLE -
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 # 

#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")

#include <bits/stdc++.h>
#include <atcoder/all>

using namespace std;
using namespace atcoder;

#define overload4(_1, _2, _3, _4, name, ...) name
#define rep1(n) for(int i = 0; i < (int)(n); ++i)
#define rep2(i, n) for(int i = 0; i < (int)(n); ++i)
#define rep3(i, a, b) for(int i = (a); i < (int)(b); ++i)
#define rep4(i, a, b, c) for(int i = (a); i < (int)(b); i += (c))
#define rep(...) overload4(__VA_ARGS__, rep4, rep3, rep2, rep1)(__VA_ARGS__)

#define rrep(i,n) for(int i = (int)(n) - 1; i >= 0; --i)
#define ALL(a) a.begin(), a.end()
#define Sort(a) sort(a.begin(), a.end())
#define RSort(a) sort(a.rbegin(), a.rend())

typedef long long int ll;
typedef unsigned long long ul;
typedef long double ld;
typedef vector<int> vi;
typedef vector<long long> vll;
typedef vector<char> vc;
typedef vector<string> vst;
typedef vector<double> vd;
typedef vector<long double> vld;
typedef pair<long long, long long> P;

template<class T> long long sum(const T& a){ return accumulate(a.begin(), a.end(), 0LL); }
template<class T> auto min(const T& a){ return *min_element(a.begin(), a.end()); }
template<class T> auto max(const T& a){ return *max_element(a.begin(), a.end()); }

const long long MINF = 0x7fffffffffff;
const long long INF = 0x1fffffffffffffff;
const long long MOD = 1000000007;
const long double EPS = 1e-9;
const long double PI = acos(-1);
 
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

template<typename T1, typename T2> istream &operator>>(istream &is, pair<T1, T2> &p){ is >> p.first >> p.second; return is; }
template<typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &p){ os << "(" << p.first << ", " << p.second << ")"; return os; }
template<typename T> istream &operator>>(istream &is, vector<T> &v){ for(T &in : v) is >> in; return is; }
template<typename T> ostream &operator<<(ostream &os, const vector<T> &v){ for(int i = 0; i < (int) v.size(); ++i){ os << v[i] << (i + 1 != (int) v.size() ? " " : ""); } return os; }
template <typename T, typename S> ostream &operator<<(ostream &os, const map<T, S> &mp){ for(auto &[key, val] : mp){ os << key << ":" << val << " "; } return os; }
template <typename T> ostream &operator<<(ostream &os, const set<T> &st){ auto itr = st.begin(); for(int i = 0; i < (int)st.size(); ++i){ os << *itr << (i + 1 != (int)st.size() ? " " : ""); itr++; } return os; }
template <typename T> ostream &operator<<(ostream &os, const multiset<T> &st){ auto itr = st.begin(); for(int i = 0; i < (int)st.size(); ++i){ os << *itr << (i + 1 != (int)st.size() ? " " : ""); itr++; } return os; }
template <typename T> ostream &operator<<(ostream &os, queue<T> q){ while(q.size()){ os << q.front() << " "; q.pop(); } return os; }
template <typename T> ostream &operator<<(ostream &os, deque<T> q){ while(q.size()){ os << q.front() << " "; q.pop_front(); } return os; }
template <typename T> ostream &operator<<(ostream &os, stack<T> st){ while(st.size()){ os << st.top() << " "; st.pop(); } return os; }
template <class T, class Container, class Compare> ostream &operator<<(ostream &os, priority_queue<T, Container, Compare> pq){ while(pq.size()){ os << pq.top() << " "; pq.pop(); } return os; }

template<class T, class U> inline T vin(T& vec, U n) { vec.resize(n); for(int i = 0; i < (int) n; ++i) cin >> vec[i]; return vec; }
template<class T> inline void vout(T vec, string s = "\n"){ for(auto x : vec) cout << x << s; }
template<class... T> void in(T&... a){ (cin >> ... >> a); }
void out(){ cout << '\n'; }
template<class T, class... Ts> void out(const T& a, const Ts&... b){ cout << a; (cout << ... << (cout << ' ', b)); cout << '\n'; }
template<class T, class U> void inGraph(vector<vector<T>>& G, U n, U m, bool directed = false){ G.resize(n); for(int i = 0; i < m; ++i){ int a, b; cin >> a >> b; a--, b--; G[a].push_back(b); if(!directed) G[b].push_back(a); } }

template <typename T>
T intpow(T x, T n){
    T ret = 1;
    while(n > 0) {
        if(n & 1) (ret *= x);
        (x *= x);
        n >>= 1;
    }
    return ret;
}

ll a;

void input(){
    in(a);
}
 
void solve(){
    ll mx = intpow(4444LL, 4LL);
    vector<vector<ll>> mod(333);
    vll pow3(4444), pow4(4444);
    for(ll i = 1; i < 4444; i++){
        pow3[i] = intpow(i, 3LL) % 333;
        pow4[i] = intpow(i, 4LL);
        mod[pow3[i] % 333].emplace_back(pow4[i]);
    }
    ll ans = 0;
    for(int i = 1; i < 4444; i++){
        for(int j = i + 1; j < 4444; j++){
            ll diff = mx - pow4[i] - pow4[j];
            if(diff < 0) break;
            ll m = a - pow3[i] - pow3[j] + 999;
            m %= 333;
            ans += (upper_bound(ALL(mod[m]), diff) - mod[m].begin()) * 2;
        }
    }
    for(int i = 1; i < 4444; i++){
        ll j = i;
        ll diff = mx - pow4[i] - pow4[j];
        ll m = a - pow3[i] - pow3[j];
        m %= 333; m += 333; m %= 333;
        ans += (upper_bound(ALL(mod[m]), diff) - mod[m].begin());
    }
    out(ans);
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    input();
    solve();
}
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