hdu-6198number 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
先写出几项找到规律是求第i个mjf−bad number就是第2*i+3项斐波那契数列的值。
矩阵快速幂模版题。之前一直没做矩阵快速幂的题，还好队友写出来了。

#include <iostream>#include <cstdio>#include <cstring>#define LL long long#define MOD 998244353using namespace std;struct matrix{    LL x[3][3];};struct matrix mutimatrix(struct matrix a,struct matrix b){    struct matrix temp;    memset(temp.x,0,sizeof(temp.x));    for(int i=0;i<2;i++)    {        for(int j=0;j<2;j++)        {            for(int k=0;k<2;k++)            {                temp.x[i][j]=temp.x[i][j]+a.x[i][k]*b.x[k][j];                temp.x[i][j]=temp.x[i][j]%MOD;            }        }    }    return temp;}struct matrix pow_matrix(struct matrix a,LL b){    struct matrix temp;    memset(temp.x,0,sizeof(temp.x));    for(int i=0;i<2;i++)    {        temp.x[i][i]=1;    }    while(b)    {        if(b&1)        {            temp=mutimatrix(temp,a);        }        a=mutimatrix(a,a);        b>>=1;    }    return temp;}int main(){    LL n;    while(scanf("%lld",&n)!=EOF)    {        struct matrix temp,ans;        temp.x[0][0]=1;        temp.x[0][1]=1;        temp.x[1][0]=1;        temp.x[1][1]=0;        ans=pow_matrix(temp,2*n+3);        printf("%lld\n",ans.x[0][1]-1);    }    return 0;}