【BM模板】_C_内存溢出

概述#include<bits/stdc++.h>#define mp make_pair#define rep(i,a,n) for (int i=a;i<n;i++)#define per(i,a,n) for (int i=n-1;i>=a;i--)#define pb push_back#define mp make_pair#define all(x) (x).beg

#include<bits/stdc++.h>#define mp make_pair#define rep(i,a,n) for (int i=a;i<n;i++)#define per(i,n) for (int i=n-1;i>=a;i--)#define pb push_back#define mp make_pair#define all(x) (x).begin(),(x).end()#define fi first#define se second#define SZ(x) ((int)(x).size())using namespace std;typedef vector<int> VI;typedef long long ll;typedef pair<int,int> PII;const ll mod = 1000000007;ll powmod(ll a,ll b) { ll res = 1; a %= mod; assert(b >= 0); for (; b; b >>= 1) { if (b & 1)res = res * a%mod; a = a * a%mod; }return res; }// headll n;namespace linear_seq {    const int N = 10010;    ll res[N],base[N],_c[N],_md[N];    vector<int> Md;    voID mul(ll *a,ll *b,int k) {        rep(i,0,k + k) _c[i] = 0;        rep(i,k) if (a[i]) rep(j,0,k) _c[i + j] = (_c[i + j] + a[i] * b[j]) % mod;        for (int i = k + k - 1; i >= k; i--) if (_c[i])            rep(j,SZ(Md)) _c[i - k + Md[j]] = (_c[i - k + Md[j]] - _c[i] * _md[Md[j]]) % mod;        rep(i,k) a[i] = _c[i];    }    int solve(ll n,VI a,VI b) { // a 系数 b 初值 b[n+1]=a[0]*b[n]+...                                  //        printf("%d\n",SZ(b));        ll ans = 0,pnt = 0;        int k = SZ(a);        assert(SZ(a) == SZ(b));        rep(i,k) _md[k - 1 - i] = -a[i]; _md[k] = 1;        Md.clear();        rep(i,k) if (_md[i] != 0) Md.push_back(i);        rep(i,k) res[i] = base[i] = 0;        res[0] = 1;        while ((1ll << pnt) <= n) pnt++;        for (int p = pnt; p >= 0; p--) {            mul(res,res,k);            if ((n >> p) & 1) {                for (int i = k - 1; i >= 0; i--) res[i + 1] = res[i]; res[0] = 0;                rep(j,SZ(Md)) res[Md[j]] = (res[Md[j]] - res[k] * _md[Md[j]]) % mod;            }        }        rep(i,k) ans = (ans + res[i] * b[i]) % mod;        if (ans<0) ans += mod;        return ans;    }    VI BM(VI s) {        VI C(1,1),B(1,1);        int L = 0,m = 1,b = 1;        rep(n,0,SZ(s)) {            ll d = 0;            rep(i,L + 1) d = (d + (ll)C[i] * s[n - i]) % mod;            if (d == 0) ++m;            else if (2 * L <= n) {                VI T = C;                ll c = mod - d * powmod(b,mod - 2) % mod;                while (SZ(C)<SZ(B) + m) C.pb(0);                rep(i,SZ(B)) C[i + m] = (C[i + m] + c * B[i]) % mod;                L = n + 1 - L; B = T; b = d; m = 1;            }            else {                ll c = mod - d * powmod(b,SZ(B)) C[i + m] = (C[i + m] + c * B[i]) % mod;                ++m;            }        }        return C;    }    int gao(VI a,ll n) {        VI c = BM(a);        c.erase(c.begin());        rep(i,SZ(c)) c[i] = (mod - c[i]) % mod;        return solve(n,c,VI(a.begin(),a.begin() + SZ(c)));    }};int main() {    cin >> n;    vector<int> V;    int a[] = {3,9,20,40,85,191,426,931,2028,4444,9765,21430,46970};    for(int i = 0; i <= 12; i++) V.push_back(a[i]);    printf("%lld\n",1ll * linear_seq::gao(V,n - 1)%mod);}