sdnu1062.Fibonacci（矩阵快速幂模板）

Description

In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 2 for n ≥ 2.

Input

a single line containing n (where 0 ≤ n ≤ 100,000,000,000)

Output

print Fn mod 1000000007 in a single line.

Sample Input

99999999999

Sample Output

669753982

Hint

An alternative formula for the Fibonacci sequence is


As a reminder, matrix multiplication is associative, and the product of two 2 × 2 matrices is given by


Also, note that raising any 2 × 2 matrix to the 0th power gives the identity matrix:

矩阵快速幂 求斐波那契  模板 

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=2;
const int mod=1000000007;
struct mat
{
    ll a[N][N];
};

mat mat_mul(mat x,mat y)
{
    mat res;
    memset(res.a,0,sizeof(res.a));
    for(int i=0;i<2;i++)
    {
        for(int j=0;j<2;j++)
        {
            for(int k=0;k<2;k++)
                res.a[i][j]=(res.a[i][j]+x.a[i][k]*y.a[k][j])%mod;
        }
    }
    return res;
}
long long mat_pow(ll n)
{
    mat c,res;
    c.a[0][0]=c.a[0][1]=c.a[1][0]=1;
    c.a[1][1]=0;
    memset(res.a,0,sizeof(res.a));
    for(int i=0;i<2;i++)
        res.a[i][i]=1;
    while(n)
    {
        if(n&1)
            res=mat_mul(res,c);
        c=mat_mul(c,c);
        n=n>>1;
    }
    return res.a[1][0];
}
int main()
{
    ll n;
    while(scanf("%lld",&n)!=EOF)
    {
        long long tmp=mat_pow(n);
        cout<<tmp<<'\n';
    }
}