zoukankan      html  css  js  c++  java
  • 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=Fn1+Fn2 (n2).

    Give you an integer k, if a positive number n can be expressed by
    n=Fa1+Fa2+...+Fak where 0a1a2ak, this positive number is mjfgood. Otherwise, this positive number is mjfbad.
    Now, give you an integer k, you task is to find the minimal positive mjfbad 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. (1k109)
     

    Output
    For each case, output the minimal mjfbad number mod 998244353.
     

    Sample Input
    1
     

    Sample Output
    4
     

    Source
     

    Recommend
    liuyiding   |   We have carefully selected several similar problems for you:  6205 6204 6203 6202 6201 
     

    Statistic | Submit | Discuss | Note

    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;
        }
    }



  • 相关阅读:
    树线段hdu 4508 美素数(线段树)
    自定义context自定义Dialog之Progress(二)
    实现语言C语言简单实现五子棋
    调用博客paip.基于HTML gui界面的javascript JS实现SLEEP。。
    水印控件windows phone中,制作一个自定义的密码输入框控件,含图片,有水印,星号显示
    请求网络网络编程
    调试网页PAIP HTML的调试与分析工具
    输出流输入输入输出流
    标记协议http协议与XML书写规范及解析技术
    描述算法10673 Play with Floor and Ceil
  • 原文地址:https://www.cnblogs.com/bryce1010/p/9387252.html
Copyright © 2011-2022 走看看