結果
| 問題 | No.215 素数サイコロと合成数サイコロ (3-Hard) |
| ユーザー |
 fumofumofuni
|
| 提出日時 | 2022-09-10 09:21:34 |
| 言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
TLE
|
| 実行時間 | - |
| コード長 | 6,362 bytes |
| コンパイル時間 | 2,405 ms |
| コンパイル使用メモリ | 213,912 KB |
| 最終ジャッジ日時 | 2025-02-07 03:51:01 |
|
ジャッジサーバーID （参考情報） |
judge5 / judge2 |
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| ファイルパターン | 結果 |
|---|---|
| other | TLE * 2 |
ソースコード
#include<bits/stdc++.h>using namespace std;#define rep(i,n) for(ll i=0;i<n;i++)#define repl(i,l,r) for(ll i=(l);i<(r);i++)#define per(i,n) for(ll i=(n)-1;i>=0;i--)#define perl(i,r,l) for(ll i=r-1;i>=l;i--)#define fi first#define se second#define pb push_back#define ins insert#define pqueue(x) priority_queue<x,vector<x>,greater<x>>#define all(x) (x).begin(),(x).end()#define CST(x) cout<<fixed<<setprecision(x)#define vtpl(x,y,z) vector<tuple<x,y,z>>#define rev(x) reverse(x);using ll=long long;using vl=vector<ll>;using vvl=vector<vector<ll>>;using pl=pair<ll,ll>;using vpl=vector<pl>;using vvpl=vector<vpl>;const ll MOD=1000000007;const ll MOD9=998244353;const int inf=1e9+10;const ll INF=4e18;const ll dy[9]={0,1,0,-1,1,1,-1,-1,0};const ll dx[9]={1,0,-1,0,1,-1,1,-1,0};template<class T> inline bool chmin(T& a, T b) {if (a > b) {a = b;return true;}return false;}template<class T> inline bool chmax(T& a, T b) {if (a < b) {a = b;return true;}return false;}const int mod = MOD;const int max_n = 200005;struct mint {ll x; // typedef long long ll;mint(ll x=0):x((x%mod+mod)%mod){}mint operator-() const { return mint(-x);}mint& operator+=(const mint a) {if ((x += a.x) >= mod) x -= mod;return *this;}mint& operator-=(const mint a) {if ((x += mod-a.x) >= mod) x -= mod;return *this;}mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}mint operator+(const mint a) const { return mint(*this) += a;}mint operator-(const mint a) const { return mint(*this) -= a;}mint operator*(const mint a) const { return mint(*this) *= a;}mint pow(ll t) const {if (!t) return 1;mint a = pow(t>>1);a *= a;if (t&1) a *= *this;return a;}bool operator==(const mint &p) const { return x == p.x; }bool operator!=(const mint &p) const { return x != p.x; }// for prime modmint inv() const { return pow(mod-2);}mint& operator/=(const mint a) { return *this *= a.inv();}mint operator/(const mint a) const { return mint(*this) /= a;}};istream& operator>>(istream& is, mint& a) { return is >> a.x;}ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}using vm=vector<mint>;using vvm=vector<vm>;struct combination {vector<mint> fact, ifact;combination(int n):fact(n+1),ifact(n+1) {assert(n < mod);fact[0] = 1;for (int i = 1; i <= n; ++i) fact[i] = fact[i-1]*i;ifact[n] = fact[n].inv();for (int i = n; i >= 1; --i) ifact[i-1] = ifact[i]*i;}mint operator()(int n, int k) {if (k < 0 || k > n) return 0;return fact[n]*ifact[k]*ifact[n-k];}}comb(max_n);vm conv(vm a,vm b){vm c(a.size()+b.size()-1);rep(i,a.size()){rep(j,b.size()){c[i+j]+=a[i]*b[j];}}return c;}vector<mint> BerlekampMassey(const vector<mint> &s) {const int N = (int)s.size();vector<mint> b, c;b.reserve(N + 1);c.reserve(N + 1);b.push_back(mint(1));c.push_back(mint(1));mint y = mint(1);for (int ed = 1; ed <= N; ed++) {int l = int(c.size()), m = int(b.size());mint x = 0;for (int i = 0; i < l; i++) x += c[i] * s[ed - l + i];b.emplace_back(mint(0));m++;if (x == mint(0)) continue;mint freq = x / y;if (l < m) {auto tmp = c;c.insert(begin(c), m - l, mint(0));for (int i = 0; i < m; i++) c[m - 1 - i] -= freq * b[m - 1 - i];b = tmp;y = x;} else {for (int i = 0; i < m; i++) c[l - 1 - i] -= freq * b[m - 1 - i];}}reverse(begin(c), end(c));return c;}template <typename mint>vector<mint> kitamasa(vector<mint> Q,vector<mint> a) {assert(!Q.empty() && Q[0] != 0);assert((int)a.size() >= int(Q.size()) - 1);vector<mint> P(Q.size()*2-2);for(ll i=0;i<Q.size()-1;i++){for(ll j=0;j<Q.size();j++){P[i+j]+=a[i]*Q[j];}}P.resize(Q.size() - 1);return P;}template<class T>struct bostan_mori {vector<T> p, q;bostan_mori(vector<T> &_p, vector<T> &_q) : p(_p), q(_q) {}void rever(vector<T> &f) const {int d = f.size();rep(i, d) if (i&1) f[i] = -f[i];}void even(vector<T> &f) const {int d = (f.size() + 1) >> 1;rep(i, d) f[i] = f[i<<1];f.resize(d);}void odd(vector<T> &f) const {int d = f.size() >> 1;rep(i, d) f[i] = f[i<<1|1];f.resize(d);}vector<T> convolution(vector<T> a,vector<T> b) const{int n=a.size(),m=b.size();vector<T> c(n+m-1);rep(i,n)rep(j,m)c[i+j]+=a[i]*b[j];return c;}T operator[] (ll n) const {vector<T> _p(p), _q(q), _q_rev(q);rever(_q_rev);for (; n; n >>= 1) {_p = convolution(move(_p), _q_rev);if (n&1) odd(_p);else even(_p);_q = convolution(move(_q), move(_q_rev));even(_q);_q_rev = _q; rever(_q_rev);}return _p[0] / _q[0];}};//https://nyaannyaan.github.io/library/fps/kitamasa.hpp//https://atcoder.jp/contests/tdpc/submissions/34362182//線形漸化式のprefixからn項目を復元できる。bostan_mori<mint> interpolation(vm a){auto q=BerlekampMassey(a);auto p=kitamasa(q,a);return bostan_mori<mint>(p,q);}int main(){ll n,p,c;cin >> n >> p >> c;vl prime={2,3,5,7,11,13};vl compos={4,6,8,9,10,12};vvm dp(301,vm(4000));dp[0][0]=1;{rep(i,6){rep(j,300){rep(k,4000){if(k+prime[i]<4000)dp[j+1][k+prime[i]]+=dp[j][k];}}}}vvm ndp(301,vm(4000));ndp[0][0]=1;{rep(i,6){rep(j,300){rep(k,4000){if(k+compos[i]<4000)ndp[j+1][k+compos[i]]+=ndp[j][k];}}}}auto f=dp[p];auto g=ndp[c];f=conv(f,g);while(f.back().x==0)f.pop_back();{ll m=f.size()*2;vm naive(m);rep(i,f.size())naive[i]=1;for(ll j=f.size();j<m;j++){rep(k,f.size()){naive[j]+=naive[j-k]*f[k];}}//rep(i,m)cout << naive[i] <<" ";cout << endl;//rep(_,f.size())naive.pop_back();auto bm=interpolation(naive);//rep(i,m)cout << bm[i] <<" ";cout << endl;cout << bm[n+f.size()-1] << endl;}}