Youhane Generator -Real Model-

C++14(g++5.4) 解法, 执行用时: 2883ms, 内存消耗: 616K, 提交时间: 2020-05-01 11:34:30

#include <bits/stdc++.h>
using namespace std;
#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).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
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;}
int _,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,0,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,0,SZ(Md)) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;
    rep(i,0,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,0,k) _md[k-1-i]=-a[i];_md[k]=1;
    Md.clear();
    rep(i,0,k) if (_md[i]!=0) Md.push_back(i);
    rep(i,0,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,0,SZ(Md)) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod;
      }
    }
    rep(i,0,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,0,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,0,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,mod-2)%mod;
      while (SZ(C)<SZ(B)+m) C.pb(0);
      rep(i,0,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,0,SZ(c)) c[i]=(mod-c[i])%mod;
    return solve(n,c,VI(a.begin(),a.begin()+SZ(c)));
  }
};
ll kv[1100];
int jc[5100];
int main() {
  int T;
  ll n,k;
  vector<int>v;
  jc[0]=jc[1]=1;
  for(int i=2;i<=5000;i++)jc[i]=1LL*i*jc[i-1]%mod;
  while (scanf("%lld%lld",&n,&k)==2) {
    v.clear();
    for(int i=0;i<k;i++)scanf("%lld",&kv[i]);
    for(int i=0;i<=min(5000LL,n);i++){
      if(i<k)v.push_back(1);
      else{
        int res=1LL*jc[i]*powmod(jc[i-k],mod-2)%mod*powmod(jc[k],mod-2)%mod;

        for(int j=0;j<k;j++){
          res=(res+1LL*kv[j]*v[i-1-j]%mod)%mod;
          //printf(",res=%d,%lld\n",v[i-1-j],kv[i]);
        }
        v.push_back(res);

      }
    }
    printf("%d\n",linear_seq::gao(v,n));
  }
}

C++11(clang++ 3.9) 解法, 执行用时: 3103ms, 内存消耗: 604K, 提交时间: 2019-03-13 01:49:14

#include<bits/stdc++.h>
using namespace std;
const int N = 4e3 + 7;
typedef long long ll;
const int mod = 1e9 + 7;

ll F[N], invF[N];
ll C(int n, int m) {
    if(n < m || m < 0) return 0;
    return F[n] * invF[m] % mod * invF[n - m] % mod;
}
ll power_mod(ll a, ll b) {
    ll ret = 1;
    for(a %= mod; b; b >>= 1, a = a * a % mod)
        if(b & 1) ret = ret * a % mod;
    return ret;
}
int n, k;
ll a[N], b[N], c[N], _c[N], res[N];

void mul(ll *a, ll *b, ll *md, int L) {
    for(int i = 0; i < 2 * L; i++) _c[i] = 0;
    for(int i = 0; i < L; i++)
        for(int j = 0; j < L; j++)
            (_c[i + j] += a[i] * b[j]) %= mod;
    for(int i = 2 * L - 2; i >= L; i--)
        for(int j = 0; j < L; j++)
            (_c[i - j - 1] += _c[i] * md[j]) %= mod;
    for(int i = 0; i < L; i++) a[i] = _c[i];
}

ll solve(ll *res, ll *r, int n, int L) {
    int pn = 0; ll ret = 0;
    while((1<<pn) <= n) pn++; pn--;
    for(int i = 0; i < L; i++) res[i] = 0;
    res[0] = 1;
    for(int i = pn; ~i; i--) {
        mul(res, res, b, L);
        if(n>>i&1) {
            for(int i = L - 1; ~i; i--)
                res[i + 1] = res[i];
            res[0] = 0;
            for(int i = 0; i < L; i++)
                (res[L - 1 - i] += res[L] * b[i]) %= mod;
        }
    }
    for(int i = 0; i < L; i++)
        (ret += r[i] * res[i]) %= mod;
    return ret;
}


int main() {
#ifdef local
    freopen("in.txt", "r", stdin);
#endif
    F[0] = invF[0] = 1;
    for(int i = 1; i < N; i++)
        F[i] = F[i - 1] * i % mod;
    invF[N - 1] = power_mod(F[N - 1], mod - 2);
    for(int i = N - 2; i; i--)
        invF[i] = invF[i + 1] * (i + 1) % mod;

    while(~scanf("%d%d", &n, &k)) {
        for(int i = 1; i <= k; i++) {
            scanf("%lld", a + i);
            if(a[i] < 0) a[i] += mod;
        }
        int m = 2 * k + 1;
        for(int i = 0; i <= m; i++) b[i] = 0;
        int sign = 1;
        for(int i = k - 1; ~i; i--, sign = mod - sign)
            b[i + k] = sign * C(k, i) % mod;
        sign = 1;
        for(int i = k; ~i; i--, sign = mod - sign) {
            ll tt = sign * C(k, i) % mod;
            for(int j = 1; j <= k; j++)
                (b[i + k - j] += tt * a[j]) %= mod;
        }
        b[m - 1] = -1;
        for(int i = m - 1; i; i--)
            b[i] = (b[i - 1] + mod - b[i]) % mod;
        b[0] = (mod - b[0]) % mod;;
        reverse(b, b + m);
        for(int i = 0; i < m; i++) {
            if(i < k) c[i] = 1;
            else {
                c[i] = C(i, k);
                for(int j = 1; j <= k; j++)
                    (c[i] += c[i - j] * a[j]) %= mod;
            }
        }
        printf("%lld\n", solve(res, c, n, m));
    }
    return 0;
}

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