ICPC2017網絡賽（瀋陽）number number number

number number number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 99    Accepted Submission(s): 64

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

 

Source

2017 ACM/ICPC Asia Regional Shenyang Online

 

Recommend

liuyiding

 

矩陣快速冪，通過計算髮現前幾組數據是4，12，33，88等

而Fn數列前幾項爲 0，1，1，2，3，5，8，13，21，34，55，89............發現規律F[2*n+4]-1  就是我們要找的公式

代碼如下：

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
typedef long long ll;
struct Matrix{
    ll matrix[5][5];
};
int n;//矩陣的階數
const int mod=998244353;
void init(Matrix &res)
{
    memset(res.matrix,0,sizeof(res.matrix));
    for(int i=0;i<n;i++)
        res.matrix[i][i]=1;
}
Matrix multiplicative(Matrix a,Matrix b)
{
    Matrix res;
    for(int i = 0 ; i < n ; i++){
        for(int j = 0 ; j < n ; j++){
            res.matrix[i][j]=0;
            for(int k = 0 ; k < n ; k++){
                res.matrix[i][j] = (res.matrix[i][j]%mod+a.matrix[i][k]*b.matrix[k][j]%mod)%mod;
            }
        }
    }
    return res;
}
Matrix pow(Matrix mx,ll m)
{
    Matrix res,base=mx;
    init(res);
    while(m)
    {
        if(m&1)
            res=multiplicative(res,base);
        base=multiplicative(base,base);
        m>>=1;
    }
    return res;
}
int main()
{
    Matrix mx,nx;
    ll m,ans;
    ll k;
    while(~scanf("%lld",&k))
    {
        n=2;
        m=2*k+4;
        mx.matrix[0][0]=mx.matrix[0][1]=mx.matrix[1][0]=1;
        mx.matrix[1][1]=0;
        nx=pow(mx,m-1);
        ans=(nx.matrix[0][0]*0+nx.matrix[0][1]*1)%mod;
        if(ans==0)
            printf("%lld\n",mod-1);
        else
            printf("%lld\n",ans-1);
    }

    return 0;
}