*递推 - 矩阵快速幂解斐波拉契数

题目：number number number

Problem Description

We define a sequence F:
F0=0,F1=1;
Fn=Fn-1+Fn-2 (n≥2).
Give you an integer k, if a positive number n can be expressed by
n=Fa1+Fa2+…+Fak where 0≤a1≤a2≤……≤ak, this positive number is mjf?good. Otherwise, this positive number is mjf?bad.
Now, give you an integer k, you task is to find the minimal positive mjf?bad number.
The answer may be too large. Please print the answer modulo 998244353.

Input

There are about 500 test cases, end up with EOF.
Each test case includes an integer k which is described above. (1≤k≤109)

Output

For each case, output the minimal mjf?bad number mod 998244353.

Sample Input

1

Sample Output

4

矩阵快速幂解斐波拉契数

矩阵快速幂与快速幂思想相同。
分析斐波拉契数论，f[i] = 1*f[i-1]+1*f[i-2]。

由f[i] = 1*f[i-1]+1*f[i-2]和f[i-1] = 1*f[i-1] + 0*f[i-2];我们可以给出矩阵形式。

所以，递推矩阵有

注：这里完全可以不用vector，用LL a[2][2]即可，vector可能超时。

#include
#include
#include
using namespace std;

typedef long long LL;
const LL mod=998244353;
typedef vector vec;
typedef vector mat;

mat mul(mat &a,mat &b)
{
    mat c(a.size(),vec(b[0].size()));
    for(int i=0; i<2; i++)
    {
        for(int j=0; j<2; j++)
        {
            for(int k=0; k<2; k++)
            {
                c[i][j]+=a[i][k]*b[k][j];
                c[i][j]%=mod;
            }
        }
    }
    return c;
}
mat pow(mat a,LL n)
{
    mat res(a.size(),vec(a.size()));
    for(int i=0; i1;//单位矩阵;
    while(n)
    {
        if(n&1) res=mul(res,a);
        a=mul(a,a);
        n/=2;
    }
    return res;
}
LL fib(LL n)
{
    mat a(2,vec(2));
    a[0][0]=1;
    a[0][1]=1;
    a[1][0]=1;
    a[1][1]=0;
    a=pow(a,n);
    return a[0][1];//也可以是a[1][0];
}
int  main()
{

    LL n;
    while(scanf("%lld",&n)==1)
    {
        printf("%lld\n",fib(2*n+3)-1);
    }
    return 0;
}