傳送門（其實就是求斐波那契數列....）

累了

明天再解釋

#include<cstdio>
#define ll long long
using namespace std;
const ll mod = 1000000007;

struct matrix {
    ll ma[2][2];
};

matrix ans;
ll n;

matrix mul(matrix A,matrix B) {
    matrix C;
    C.ma[0][0] = C.ma[0][1] = C.ma[1][0] = C.ma[1][1] = 0;
    for(int i = 0
 
; i < 2; i++)
        for(int j = 0; j < 2; j++)
            for(int k = 0; k < 2; k++)
                C.ma[i][j] += A.ma[i][k] * B.ma[k][j] % mod;
    return C;
}

matrix quickpow(matrix A,ll n) {
    matrix B;
    B.ma[0][0] = B.ma[1][1] = 1;
    B.ma[0][1] = B.ma[1][0] = 0;
    while(n) {
        
 
if(n&1)B = mul(A,B);
        A = mul(A,A);
        n >>= 1;
    }
    return B;
}

int main() {
    scanf("%lld",&n);
    matrix A;
    A.ma[0][0] = A.ma[0][1] = A.ma[1][0] = 1;
    A.ma[1][1] = 0;
    ans = quickpow (A,n);
    printf("%lld",ans.ma[0][1]%mod);
    return 0;
}