HDU6198 number number number

We define a sequence F:
⋅⋅ $F_{0}=0, F_{1}=1$
⋅⋅ $F_{n}=F_{n−1}+F_{n−2} (n≥2).$

Give you an integer $k$, if a positive number $n$ can be expressed by
$n=F_{a_{1}}+F_{a_{2}}+...+F_{a_{k}}$ where $0≤a_{1}≤a_{2}≤⋯≤a_{k}$, 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 kk which is described above. $(1≤k≤10^9)$

Output
For each case, output the minimal mjf−badmjf−bad number mod 998244353.

Sample Input
1

Sample Output
4

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结论：求$fib_{0}(2n+3)$。$O(n)$的做法会超时，所以需要$O(logn)$的做法。
过程：暂时将「用一系列fibonacci数的和表示」这一操作称作「fib展开」，所用fibonacci数的数目最小的展开称作「最小展开」；题目要求的是「最小的·最小fib展开中fib数的数目大于k的数」。易知任意最小展开都不可能包含类似于$fib_{0}(n) + fib_{0}(n+1)$这样的东西，即所用的fib数一定是非相邻的；要求最小，所以符合要求的数的最小fib展开中任意两个fib数之间只会隔一个fib数。于是有两种可能：
1. $fib_{0}(0)+fib_{0}(2)+fib_{0}(4)+...$
2. $fib_{0}(1)+fib_{0}(3)+fib_{0}(5)+...$
但是第二种是不可能的。将第一种通过$fib_{0}(n)=fib_{0}(n-1)+fib_{0}(n-2)$展开得到$fib_{0}(0)+fib_{0}(1)+fib_{0}(2)+...$，即$fib_{0}$前n项和；设为数列$S_{fib_{0}}$；考察可知这个数列的最小展开的fib数的数目为1 1 1 2 2 3 3 4 4 5 5… 猜测结果为$S_{fib_{0}}(2k+1)$，再次观察数列猜测$S_{fib_{0}}(2k+1)=fib_{0}(2k+3)-1$。
接下来就只剩下求$fib_{0}$的问题了；500个testcase，k最大可以有$10^9$，可以知道用$O(n)$的做法一定会TLE。网上面找到的做法都涉及矩阵快速幂；下面是一个来自这里的更加简单的做法：
$fib_{0}(2k) = fib_{0}(k)[2fib_{0}(k+1)-fib_{0}(k)$
$fib_{0}(2k+1) = fib_{0}(k+1)^2 + fib_{0}(k)^2$
变换一下就是：
$fib_{0}(k) = fib_{0}(\frac{k}{2})[2fib_{0}(\frac{k}{2}+1)-fib_{0}(\frac{k}{2})]$ (k is even)
$fib_{0}(k) = fib_{0}(\frac{k+1}{2})^2+fib_{0}(\frac{k-1}{2})^2$ (k is odd)
然后直接递归就好了。但是不能直接按公式做两次调用，那样会退化到会TLE的程度。
推理过程可以说是非常暴力了。也许好好证明之后结果也是这样吧。

#include<stdio.h>
#include<string.h>

typedef long long ll;

const int MOD = 998244353;
int K;

void quickfibx(int k, ll& a, ll& b){
    if(k==0){ a = 0, b = 1; return ; }
    if(k==1){ a = 1, b = 1; return ; }
    if(k==2){ a = 1, b = 2; return ; }
    quickfibx((k>>1),a,b);
    // now a = F(k>>1), b = F(k>>1 + 1);
    ll c = (a*((((2*b)%MOD)-a)%MOD))%MOD,
       d = (((a*a)%MOD)+((b*b)%MOD))%MOD;
    if(k%2==0){ a = c, b = d; }
    else { a = d, b = (c+d)%MOD; }
}
ll quickfib(int k){
    ll a, b;
    quickfibx(k,a,b);
    return a;
}

int main(){
    while(~scanf("%d",&K)){
        printf("%d\n",quickfib((K<<1)+3)-1);
    }
    return 0;
}