hdu6198 number number number（递推公式黑科技）

number number number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 192    Accepted Submission(s): 126

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

 

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题意：斐波拉契数列，求不能由这些k个斐波那契数列数组成的最小整数

思路：先手写找规律，再用黑科技代码模板

//递推公式黑科技
#include<bits/stdc++.h>
using namespace std;
#define X first
#define Y second
#define PB push_back
#define MP make_pair
#define MEM(x,y) memset(x,y,sizeof(x));
#define bug(x) cout<<"bug"<<x<<endl;
typedef long long ll;
typedef pair<int,int> pii;
using namespace std;
const int maxn=1e3+10;
const int mod=998244353;
ll powmod(ll a,ll b){
    ll res=1;a%=mod;
    assert(b>=0);
    for(;b;b>>=1){
        if(b&1)res=res*a%mod;a=a*a%mod;
    }
    return res;
}
// head
namespace linear_seq {
    const int N=10010;
    ll res[N],base[N],_c[N],_md[N];
    vector<int> Md;
    void mul(ll *a,ll *b,int k) {
        for(int i=0;i<k+k;i++) _c[i]=0;
        for(int i=0;i<k;i++)
            if (a[i])
                for(int j=0;j<k;j++) _c[i+j]=(_c[i+j]+a[i]*b[j])%mod;
        for (int i=k+k-1;i>=k;i--)
            if (_c[i])
                for(int j=0;j<Md.size();j++)
                    _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;
        for(int i=0;i<k;i++) a[i]=_c[i];
    }
    int solve(ll n,vector<int> a,vector<int> b) {
    // a 系数 b 初值 b[n+1]=a[0]*b[n]+...
        ll ans=0,pnt=0;
        int k=a.size();
        assert(a.size()==b.size());
        for(int i=0;i<k;i++) _md[k-1-i]=-a[i];_md[k]=1;
        Md.clear();
        for(int i=0;i<k;i++) if (_md[i]!=0) Md.push_back(i);
        for(int i=0;i<k;i++) res[i]=base[i]=0;
        res[0]=1;
        while ((1ll<<pnt)<=n) pnt++;
        for (int p=pnt;p>=0;p--) {
            mul(res,res,k);
            if ((n>>p)&1) {
                for (int i=k-1;i>=0;i--) res[i+1]=res[i];res[0]=0;
                for(int j=0;j<Md.size();j++) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod;
            }
        }
        for(int i=0;i<k;i++) ans=(ans+res[i]*b[i])%mod;
        if (ans<0) ans+=mod;
        return ans;
    }
    vector<int> BM(vector<int> s) {
        vector<int> C(1,1),B(1,1);
        int L=0,m=1,b=1;
        for(int n=0;n<s.size();n++) {
            ll d=0;
            for(int i=0;i<L+1;i++) d=(d+(ll)C[i]*s[n-i])%mod;
            if (d==0) ++m;
            else if (2*L<=n) {
                vector<int> T=C;
                ll c=mod-d*powmod(b,mod-2)%mod;
                while (C.size()<B.size()+m) C.PB(0);
                for(int i=0;i<B.size();i++) C[i+m]=(C[i+m]+c*B[i])%mod;
                L=n+1-L; B=T; b=d; m=1;
            } else {
                ll c=mod-d*powmod(b,mod-2)%mod;
                while (C.size()<B.size()+m) C.PB(0);
                for(int i=0;i<B.size();i++) C[i+m]=(C[i+m]+c*B[i])%mod;
                ++m;
            }
        }
        return C;
    }
    int gao(vector<int> a,ll n) {
        vector<int> c=BM(a);
        c.erase(c.begin());
        for(int i=0;i<c.size();i++) c[i]=(mod-c[i])%mod;
        return solve(n,c,vector<int>(a.begin(),a.begin()+c.size()));
    }
};

int main(){
    ll t,n;
//    cin>>t;
    while(cin>>n){
        cout<<(linear_seq::gao(vector<int>{5,13,34,89},n-1)%mod-1)%mod<<endl;
    }
}