LOJ 6291 小L进阶的斐波那契数列游戏

题目链接https://loj.ac/problem/6291

题意:略。

思路:观察前几项发现f_1^2+f_2^2+...+f_n^2=f_nf_{n+1},直接上矩阵快速幂就可以了,或者用黑科技连规律都不用找了。

矩阵快速幂

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int G=2;
const int P=1000000007;
struct matrix{
    #define MS(x,y) memset(x,y,sizeof(x))
    #define MC(x,y) memcpy(x,y,sizeof(x))
    int v[G][G];
    void O(){MS(v,0);}
    void E(){MS(v,0);for(int i=0;i<G;++i)v[i][i]=1;}
    void print(){
        for (int i=0;i<G;i++){
            for (int j=0;j<G;j++){
                printf("%d ",v[i][j]);
            }
            puts("");
        }
    }
    matrix operator*(const matrix &b)const{
        matrix c;c.O();
        for (int k=0;k<G;k++){
            for (int i=0;i<G;i++)if(v[i][k]){
                for (int j=0;j<G;j++){
                    c.v[i][j]=(c.v[i][j]+(ll)v[i][k]*b.v[k][j])%P;
                }
            }
        }
        return c;
    }
    matrix operator+(const matrix &b)const{
        matrix c;c.O();
        for (int i=0;i<G;i++){
            for (int j=0;j<G;j++){
                c.v[i][j]=(v[i][j]+b.v[i][j])%P;
            }
        }
        return c;
    }
    matrix operator^(ll p)const{
        matrix y; y.E();
        matrix x; memcpy(x.v,v,sizeof(v));
        while (p){
            if (p&1) y=y*x;
            x=x*x;
            p>>=1;
        }
        return y;
    }
};
ll n;
int main(){
    scanf("%lld",&n);
    matrix res;
    res.v[0][0]=res.v[0][1]=res.v[1][0]=1,res.v[1][1]=0;
    res=res^(n-1);
    int a=(res.v[0][0]+res.v[0][1])%P;
    int b=(res.v[1][0]+res.v[1][1])%P;
    printf("%d\n",1LL*a*b%P);
    return 0;
}

黑科技

#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
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;}
ll 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) {
        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)));
    }
};
int main() {
    scanf("%lld",&n);
    printf("%d\n",linear_seq::gao(VI{1,2,6,15,40,104},n-1));
}

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