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

Source
2017 ACM/ICPC Asia Regional Shenyang Online

程式碼

#include<bits/stdc++.h>
 

using namespace std ;
typedef long long LL ;

const int MAXN = 100000+10;
const int MAXM = 1e5 ;
const LL mod  =  998244353;
const int  inf = 0x3f3f3f3f; 

struct Matirx {
    int h,w;
    LL a[5][5];
}ori,res,it;
LL f[5]={0,1,1};
void init(){
    it.w=2;it.h=1;it.a[1][1]=1;it.a[1][2]=0;
    res.w=res.h=2;
    memset
 
(res.a,0,sizeof(res.a));
    res.a[1][1]=res.a[2][2]=1;
    ori.w=ori.h=2;
    memset(ori.a,0,sizeof(ori.a));
    ori.a[1][1]= 1;ori.a[1][2]= 1;
    ori.a[2][1]= 1;ori.a[2][2]= 0;  
}
Matirx multy(Matirx x,Matirx y){
    Matirx z;z.w=y.w;z.h=x.h;
    memset(z.a,0,sizeof(z.a));
    for(int i=1;i<=x.h;i++){
        for(int k=1;k<=x.w;k++){
            if(x.a[i][k]==0) continue;
            for(int j=1;j<=y.w;j++)
                z.a[i][j]=(z.a[i][j]+x.a[i][k]*y.a[k][j]%mod)%mod;
        }
    }
    return z;
}

LL Matirx_mod(LL n){
    if(n<2) return f[n];
    else n-=1;
    while(n){
        if(n&1) res=multy(ori,res);
        ori=multy(ori,ori);
        n>>=1;
    }
    res=multy(it,res);
    return res.a[1][1]%mod;
}
int main(){
    LL k;
    while(~scanf("%lld",&k)){
        init(); 
        printf("%lld\n",(Matirx_mod(2*k+3)%mod-1+mod)%mod); 
     }
    return  0;
}