We define a sequence F:
⋅⋅
⋅⋅
Give you an integer
Now, give you an integer
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.
Output
For each case, output the minimal mjf−badmjf−bad number mod 998244353.
Sample Input
1
Sample Output
4
结论:求
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
1.
2.
但是第二种是不可能的。将第一种通过
接下来就只剩下求
变换一下就是:
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;
}