2017 ACM/ICPC Asia Regional Shenyang Online：number number number

number number number

                                                                   Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                                                Total Submission(s): 0    Accepted Submission(s): 0

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

思路：找规律，当k=1时，n=F5-1=4。k=2，n=F7-1=12。k=3，n=F9-1=33。所以大胆推测n=F(2*k+3)-1；即矩阵快速幂。

#include<bits/stdc++.h>using namespace std;const int MOD=998244353;struct lenka{    long long a[2][2];};lenka cla(const lenka& a,const lenka& b){    lenka c;    memset(c.a,0,sizeof c.a);    for(int i=0;i<2;i++)    {        for(int j=0;j<2;j++)        {            for(int k=0;k<2;k++)            {                c.a[i][j]+=a.a[i][k]*b.a[k][j];                c.a[i][j]%=MOD;            }        }    }    return c;}long long POW(int k){    lenka res,a;    memset(res.a,0,sizeof res.a);    memset(a.a,0,sizeof a.a);    for(int i=0;i<2;i++)res.a[i][i]=1;    a.a[0][0]=1,a.a[0][1]=1;    a.a[1][0]=1,a.a[1][1]=0;    while(k)    {        if(k&1)res=cla(res,a);        a=cla(a,a);        k/=2;    }    return (res.a[0][0]+res.a[1][0])%MOD;}int main(){    long long k;    while(scanf("%lld",&k)!=EOF)printf("%lld\n",(POW(2*k+1)-1+MOD)%MOD);    return 0;}