HDU6198 number number number

题意：

在斐波那契求k个数的和，求他们的和不能形成的最小整数。例如：0 1 1 2 3 5 8 13 21 34 55 89...求k＝1不能形成的最小的整数，显然是4，k＝2时，1+1=2，0+1=1，1+2=3，1＋3=4，2+3=5，1+5=6，2+5=7，3＋5=8，1+8=9，2+8=10，3+8=11，5+8=13，...可以看出没有12，所以k＝2时的和不能形成12.发现规律是第5.7.9.11....   -1，得到0 1 1 2 4 8 12 21 33 55 88.....  因此得出规律是f(3+2*k).

由于数据太大所以用斐波那契的矩阵快速幂求。

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

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

#include <iostream>#include <cstddef>#include <cstring>#include <vector>using namespace std;typedef long long ll;const int mod=998244353;typedef vector<ll> vec;typedef vector<vec> mat;mat mul(mat &a,mat &b)//表示不会这样用，，，，{    mat c(a.size(),vec(b[0].size()));    for(int i=0; i<2; i++)    {        for(int j=0; j<2; j++)        {            for(int k=0; k<2; k++)            {                c[i][j]+=a[i][k]*b[k][j];                c[i][j]%=mod;            }        }    }    return c;}mat pow(mat a,ll n){    mat res(a.size(),vec(a.size()));    for(int i=0; i<a.size(); i++)        res[i][i]=1;//单位矩阵；    while(n)    {        if(n&1) res=mul(res,a);        a=mul(a,a);        n/=2;    }    return res;}ll solve(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);    return a[0][1]%998244353;//也可以是a[1][0];}int main(){    ll k;    while(~scanf("%lld",&k))    {        cout<<solve(3+2*k)%998244353-1<<endl;    }    return 0;}