number number number HDU - 6198(矩阵快速幂)

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
#include <cstdio>
#include <string>
#include <string.h>
#include <iostream>
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long LL;
const LL MOD = 998244353;
struct node
{
    LL m[2][2];
} ans,base;
node multi(node a,node b)
{
    node tmp;
    mem(tmp.m,0);
    for(int i=0; i<2; i++)
        for(int j=0; j<2; j++)
            for(int k=0; k<2; k++)
                tmp.m[i][j]=(tmp.m[i][j]+a.m[i][k]*b.m[k][j]%MOD)%MOD;
    return tmp;
}
LL fast_mod(LL n)
{
  base.m[0][0]=base.m[0][1]=base.m[1][0]=1;
  base.m[1][1]=0;
  ans.m[0][0]=ans.m[1][1]=1;
  ans.m[0][1]=ans.m[1][0]=0;
  while(n)
  {
     if(n&1)
     {
        ans=multi(ans,base);
     }
     base=multi(base,base);
     n>>=1;
  }
  return ans.m[0][1];
}
int main()
{
    LL n;
    while(~scanf("%lld",&n))
    {
       printf("%lld\n",fast_mod(n*2+3)-1);
    }
    return 0;
}
### HDU 1466 Problem Description and Solution The problem **HDU 1466** involves calculating the expected number of steps to reach a certain state under specific conditions. The key elements include: #### Problem Statement Given an interactive scenario where operations are performed on numbers modulo \(998244353\), one must determine the expected number of steps required to achieve a particular outcome. For this type of problem, dynamic programming (DP) is often employed as it allows breaking down complex problems into simpler subproblems that can be solved iteratively or recursively with memoization techniques[^1]. In more detail, consider the constraints provided by similar problems such as those found in references like HDU 6327 which deals with random sequences using DP within given bounds \((1 \leq T \leq 10, 4 \leq n \leq 100)\)[^2]. These types of constraints suggest iterative approaches over small ranges might work efficiently here too. Additionally, when dealing with large inputs up to \(2 \times 10^7\) as seen in reference materials related to counting algorithms [^4], efficient data structures and optimization strategies become crucial for performance reasons. However, directly applying these methods requires understanding how they fit specifically into solving the expectation value calculation involved in HDU 1466. For instance, if each step has multiple outcomes weighted differently based on probabilities, then summing products of probability times cost across all possible states until convergence gives us our answer. To implement this approach effectively: ```python MOD = 998244353 def solve_expectation(n): dp = [0] * (n + 1) # Base case initialization depending upon problem specifics for i in range(1, n + 1): total_prob = 0 # Calculate transition probabilities from previous states for j in transitions_from(i): # Placeholder function representing valid moves prob = calculate_probability(j) next_state_cost = get_next_state_cost(j) dp[i] += prob * (next_state_cost + dp[j]) % MOD total_prob += prob dp[i] %= MOD # Normalize current state's expectation due to accumulated probability mass if total_prob != 0: dp[i] *= pow(total_prob, MOD - 2, MOD) dp[i] %= MOD return dp[n] # Example usage would depend heavily on exact rules governing transitions between states. ``` This code snippet outlines a generic framework tailored towards computing expectations via dynamic programming while adhering strictly to modular arithmetic requirements specified by the contest question format.
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