HDU

number number number

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

 

解题思路：矩阵快速幂求斐波那契数列，队友A了

#include<iostream>#include<deque>#include<memory.h>#include<stdio.h>#include<map>#include<string>#include<algorithm>#include<vector>#include<math.h>#include<stack>#include<queue>#include<set>using namespace std;const int mod=998244353;typedef long long int ll;typedef vector<long long> vec;typedef vector<vec> mat;ll n;mat mul(mat A,mat B){    mat C(A.size(),vec(B[0].size()));    for(int i=0;i<A.size();i++)    {        for(int j=0;j<B.size();j++)        {            for(int k=0;k<B[0].size();k++)            {                C[i][j]+=(A[i][k]*B[k][j])%mod;                C[i][j]%=mod;            }        }    }    return C;}mat pow(mat A,ll n){    mat B(A.size(),vec(A.size()));    for(int i=0;i<A.size();i++)    {        B[i][i]=1;    }    while(n)    {        if(n&1)        {            B=mul(B,A);        }        A=mul(A,A);        n>>=1;    }    return B;}void slove(ll n){    mat A(2,vec(2));    A[0][0]=1; A[0][1]=1;    A[1][0]=1; A[1][1]=0;    A=pow(A,n);    printf("%lld\n",A[1][0]-1);}int main(){    while(scanf("%lld",&n)!=EOF)    {        slove(2*(n+1)+1);    }    return 0;}