結果

問題 No.890 移調の限られた旋法
コンテスト
ユーザー tonegawa
提出日時 2020-08-13 03:57:10
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
TLE  
実行時間 -
コード長 2,921 bytes
コンパイル時間 1,751 ms
コンパイル使用メモリ 119,432 KB
最終ジャッジ日時 2025-01-12 21:37:13
ジャッジサーバーID
（参考情報） 		judge1 / judge1
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 * 10 TLE * 22
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 <iostream>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <algorithm>
#include <set>
#include <map>
#include <bitset>
#include <cmath>
#include <functional>
#include <iomanip>
#define vll vector<ll>
#define vvvl vector<vvl>
#define vvl vector<vector<ll>>
#define VV(a, b, c, d) vector<vector<d>>(a, vector<d>(b, c))
#define VVV(a, b, c, d) vector<vvl>(a, vvl(b, vll (c, d)));
#define re(c, b) for(ll c=0;c<b;c++)
#define all(obj) (obj).begin(), (obj).end()
typedef long long int ll;
typedef long double ld;
using namespace std;

#define P 1000000007
#define N_MAX 10000000
ll fac[N_MAX+1], inv[N_MAX+1], finv[N_MAX+1];

ll comb(ll n, ll k){
  if(n<0||k<0||n<k) return 0;
  return (((fac[n]*finv[n-k])%P)*finv[k])%P;
}
ll perm(ll n, ll k){
  if(n<0||k<0||n<k) return 0;
  return (fac[n]*finv[n-k])%P;
}
void init(){
  fac[0] = finv[0] = fac[1] = finv[1] = inv[1] = 1;
  for(int i = 2; i <= N_MAX; i++){
    fac[i] = (fac[i-1]*i)%P;
    inv[i] = ((-(P/i)*inv[P%i])%P+P)%P;
    finv[i] = (finv[i-1]*inv[i])%P;
  }
}
ll mpow(ll a, ll b){
  ll ret = 1, num = a;
  while(b>0){
    if(b%2) ret = (ret*num)%P;
    num = (num*num)%P;
    b /= 2;
  }
  return ret;
}

struct ml{
  ll n;
  ml(ll n):n(n){if(n>=P) n%=P;}
  ml operator + (ml a){return (this->n + a.n)%P;}
  ml operator - (ml a){return (this->n - a.n + P)%P;}
  ml operator * (ml a){return (this->n * a.n)%P;}
  ml operator / (ml a){return (this->n*(a.n<=N_MAX?inv[a.n]:mpow(a.n,P-2)))%P;}
  ml operator + (ll a){return (this->n + a)%P;}
  ml operator - (ll a){return (this->n - a + P)%P;}
  ml operator * (ll a){return (this->n * a)%P;}
  ml operator / (ll a){return (this->n*(a<=N_MAX?inv[a]:mpow(a,P-2)))%P;}
  ml operator ^ (ll a){return mpow(this->n, a);}

  void operator += (ml a){n = (this->n + a.n)%P;}
  void operator -= (ml a){n =  (this->n - a.n + P)%P;}
  void operator *= (ml a){n =  (this->n * a.n)%P;}
  void operator /= (ml a){n = (this->n*(a.n<=N_MAX?inv[a.n]:mpow(a.n,P-2)))%P;}
  void operator += (ll a){n =  (this->n + a)%P;}
  void operator -= (ll a){n = (this->n - a + P)%P;}
  void operator *= (ll a){n =  (this->n * a)%P;}
  void operator /= (ll a){n = (this->n*(a<=N_MAX?inv[a]:mpow(a,P-2)))%P;}
  void operator ^= (ll a){n = mpow(this->n, a);}
};


int main(int argc, char const *argv[]) {
  init();
  ll n,k;std::cin >> n>>k;

  vll d;
  for(ll i=2;i*i<=n;i++){
    if(n%i==0){
      d.push_back(i);
      if(i*i!=n) d.push_back(n/i);
    }
  }
  if(n!=1) d.push_back(n);
  ll x = k, y =  n - k;

  vll ans(100001, 0);
  for(auto e:d){
    if(x%e==0&&y%e==0){
      ans[e] = comb((x+y)/e, x/e);
    }
  }
  sort(all(d));
  for(int i=d.size()-1;i>=0;i--){
    for(int j=i-1;j>=0;j--){
      if(x%d[i]||x%d[j]||y%d[i]||y%d[j]) continue;
      if(d[i]%d[j]==0) ans[d[j]] = (ans[d[j]] - ans[d[i]] + P)%P;
    }
  }
  ll s = 0;
  for(auto e:d) s = (s + ans[e])%P;
  std::cout << s << '\n';
  return 0;
}
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