matrixquickpow

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int MOD = 1e9+7;
typedef long long ll;
struct Node{
    ll a[2][2];
};

Node mul(Node a, Node b){
    Node ans;
    for(int i=0; i<2; ++i){
        for(int j=0; j<2; ++j){
            ans.a[i][j]=0;
            for(int k=0; k<2; ++k)
                ans.a[i][j]=(ans.a[i][j]+a.a[i][k]*b.a[k][j]%MOD)%MOD;
        }
    }
    return ans;
}

Node qpow(Node a,ll b){
    Node ans;
    for(int i=0; i<2; ++i)
        for(int j=0; j<2; ++j)
            ans.a[i][j]=(i==j);
    while(b){
        if(b&1)
            ans=mul(ans,a);
        b>>=1;
        a=mul(a,a);
    }
    return ans;
}

int main (){
    ll n;
    scanf("%lld",&n);
    Node base;
    int c[2][2]={1,1,1,0};
    for(int i=0; i<2; ++i)
        for(int j=0; j<2; ++j)
            base.a[i][j]=c[i][j];
    Node ans=qpow(base,n);
    printf("%lld\n",(ans.a[1][0]*ans.a[0][0])%MOD);
    return 0;
}

dujiaoBM

#include<bits/stdc++.h>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define pb push_back
typedef long long ll;
#define SZ(x) ((ll)(x).size())
typedef vector<ll> VI;
typedef pair<ll, ll> PII;
const ll mod = 1000000007;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); while (ch < 48 || ch > 57) { if (ch == '-') w = -1; ch = getchar(); }    while (ch >= 48 && ch <= 57) s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar();    return s * w; }
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; }
// head

ll _, n;
namespace linear_seq {
    const ll N = 10010;
    ll res[N], base[N], _c[N], _md[N];

    vector<ll> Md;
    void mul(ll* a, ll* b, ll 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 (ll 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];
    }
    ll 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;
        ll 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 (ll p = pnt; p >= 0; p--) {
            mul(res, res, k);
            if ((n >> p) & 1) {
                for (ll 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);
        ll 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;
    }
    ll 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() {
    //while (~scanf("%lld", &n)) {
    ll n;    scanf("%lld", &n);
    vector<ll>v;
    ll a=1,b=1,c;
    v.push_back(1);//前几项
    v.push_back(1);
    for(int i=3;i<=100;++i)
        c=(a+b)%mod,a=b,b=c,v.push_back(c);
    //输入n ,输出第n项的值  一般大于10项即可出答案，越多越好
    printf("%d\n", linear_seq::gao(v, n - 1) * linear_seq::gao(v, n) % mod);
    //}
}