結果

問題 No.890 移調の限られた旋法
ユーザー NyaanNyaanNyaanNyaan
提出日時 2019-09-20 21:52:05
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 4,274 bytes
コンパイル時間 2,084 ms
コンパイル使用メモリ 180,424 KB
実行使用メモリ 26,880 KB
最終ジャッジ日時 2024-09-14 17:17:11
合計ジャッジ時間 4,691 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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ファイルパターン 結果
sample AC * 3
other AC * 24 WA * 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 #
プレゼンテーションモードにする

#include <bits/stdc++.h>
#define whlie while
#define pb push_back
#define eb emplace_back
#define fi first
#define se second
#define rep(i,N) for(int i = 0; i < (N); i++)
#define repr(i,N) for(int i = (N) - 1; i >= 0; i--)
#define rep1(i,N) for(int i = 1; i <= (N) ; i++)
#define repr1(i,N) for(int i = (N) ; i > 0 ; i--)
#define each(x,v) for(auto& x : v)
#define all(v) (v).begin(),(v).end()
#define sz(v) ((int)(v).size())
#define vrep(v,it) for(auto it = v.begin(); it != v.end(); it++)
#define vrepr(v,it) for(auto it = v.rbegin(); it != v.rend(); it++)
#define ini(...) int __VA_ARGS__; in(__VA_ARGS__)
#define inl(...) ll __VA_ARGS__; in(__VA_ARGS__)
#define ins(...) string __VA_ARGS__; in(__VA_ARGS__)
using namespace std; void solve();
using ll = long long; using vl = vector<ll>;
using vi = vector<int>; using vvi = vector< vector<int> >;
constexpr int inf = 1001001001;
constexpr ll infLL = (1LL << 61) - 1;
struct IoSetupNya {IoSetupNya() { cin.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(15); cerr << fixed << setprecision(7
    );} } iosetupnya;
template<typename T, typename U> inline bool amin(T &x, U y) { return (y < x) ? (x = y, true) : false; }
template<typename T, typename U> inline bool amax(T &x, U y) { return (x < y) ? (x = y, true) : false; }
template<typename T, typename U> ostream& operator <<(ostream& os, const pair<T, U> &p) { os << p.first << " " << p.second; return os; }
template<typename T, typename U> istream& operator >>(istream& is, pair<T, U> &p) { is >> p.first >> p.second; return is; }
template<typename T> ostream& operator <<(ostream& os, const vector<T> &v) { int s = (int)v.size(); rep(i,s) os << (i ? " " : "") << v[i]; return os;
    }
template<typename T> istream& operator >>(istream& is, vector<T> &v) { for(auto &x : v) is >> x; return is; }
void in(){} template <typename T,class... U> void in(T &t,U &...u){ cin >> t; in(u...);}
void out(){cout << "\n";} template <typename T,class... U> void out(const T &t,const U &...u){ cout << t; if(sizeof...(u)) cout << " "; out(u...);}
template<typename T>void die(T x){out(x); exit(0);}
#ifdef NyaanDebug
#include "NyaanDebug.h"
#define trc(...) do { cerr << #__VA_ARGS__ << " = "; dbg_out(__VA_ARGS__);} while(0)
#define trca(v,...) do { cerr << #v << " = "; array_out(v , __VA_ARGS__ );} while(0)
#else
#define trc(...)
#define trca(...)
int main(){solve();}
#endif
using P = pair<ll,ll>; using vp = vector<P>;
constexpr int MOD = /**/ 1000000007; //*/ 998244353;
/////////////////
vector<ll> fac,finv,inv;
void COMinit(int MAX) {
MAX++;
fac.resize(MAX , 0);
finv.resize(MAX , 0);
inv.resize(MAX , 0);
fac[0] = fac[1] = finv[0] = finv[1] = inv[1] = 1;
for (int i = 2; i < MAX; i++){
fac[i] = fac[i - 1] * i % MOD;
inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
finv[i] = finv[i - 1] * inv[i] % MOD;
}
}
// nCk combination
inline long long COM(int n,int k){
if(n < k || k < 0 || n < 0) return 0;
else return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
// nPk permutation
inline long long PER(int n,int k){
if (n < k || k < 0 || n < 0) return 0;
else return (fac[n] * finv[n - k]) % MOD;
}
// nHk homogeneous polynomial
inline long long HGP(int n,int k){
if(n == 0 && k == 0) return 1; //?
else if(n < 1 || k < 0) return 0;
else return fac[n + k - 1] * (finv[k] * finv[n - 1] % MOD) % MOD;
}
// c++17define
#define gcd nyagcd
#define lcm nyalcm
ll nyagcd(ll x, ll y){
ll z;
if(x > y) swap(x,y);
while(x){
x = y % (z = x); y = z;
}
return y;
}
ll nyalcm(ll x,ll y){
return 1LL * x / gcd(x,y) * y;
}
void solve(){
COMinit(1001000);
ini(N,K);
ll ans = 0;
ll cur = gcd(N , K);
trc(cur);
if(cur == 1) die(0);
map<ll,int> m;
for(ll i=2;i *i <= N;i++){
whlie(cur % i == 0){
m[i]++;
cur /= i;
}
}
if(cur != 1) m[cur]++;
int x = m.size();
vi memo;
each(x,m) memo.eb(x.fi);
rep(bit , 1LL << x){
if(bit == 0) continue;
ll cur = 1;
int cnt = 0;
rep(i,x){
if(bit & (1 << i)) cur *= memo[i] , cnt++;
}
trc(cur , cnt);
if(cnt % 2 == 0) ans -= COM(x-1,cnt-1) *COM(N/cur,K/cur) % MOD;
else ans +=COM(x-1,cnt-1) * COM(N/cur , K/cur) % MOD;
}
out(ans);
}
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