zoukankan      html  css  js  c++  java
  • The Nth Item (The 2019 Asia Nanchang First Round Online Programming Contest)

    #include<cstdio>
    #include<cmath>
    #include<iostream>
    #include<cstring>
    #include<algorithm>
    #include<map>
    #include<vector>
    using namespace std;
    typedef  long long ll;
    typedef vector<ll>vec;
    typedef vector<vec>mat;
    ll M=998244353;
    map<ll,ll>cc;
    mat mul(mat &A,mat &B){
    mat C(A.size(),vec(B[0].size()));
    for(int i=0;i<A.size();i++)
        for(int k=0;k<B.size();k++)
        for(int j=0;j<B[0].size();j++)
    {
        C[i][j]=(C[i][j]+A[i][k]*B[k][j])%M;
    }
    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>0)
    {
        if(n&1) B=mul(B,A);
        A=mul(A,A);
        n>>=1;
    
    }
    return B;
    }
    map<ll,ll>kkk;
    int main()
    {
        mat A(2,vec(2));
        A[0][0]=3,A[0][1]=2,A[1][0]=1,A[1][1]=0;
        mat B(2,vec(2));
        ll q,n;
        kkk.clear();
        scanf("%lld%lld",&q,&n);
        ll dns=0,k,kk1;
        while(q--)
        {
               B=pow(A,n);
               ll kk=dns;
              // cout<<B[1][0]<<' '<<n<<endl;;
            dns^=B[1][0];
            kkk[n]=B[1][0];
            n=n^(B[1][0]*B[1][0]);
            if(dns==kk1)
            {
                if(q%2==1)
                    dns=kk;
                else
                    dns=kk1;
                break;
            }
    
                kk1=kk;
            //cout<<dns<<endl;
        }
        printf("%lld
    ",dns);
    }

    For a series FF:

    displaystyle egin{gathered} F(0) = 0,F(1) = 1\ F(n) = 3*F(n-1)+2*F(n-2),(n geq 2) end{gathered}F(0)=0,F(1)=1F(n)=3F(n1)+2F(n2),(n2)

     

    We have some queries. For each query NN, the answer AA is the value F(N)F(N) modulo 998244353998244353.

    Moreover, the input data is given in the form of encryption, only the number of queries QQ and the first query N_1N1are given. For the others, the query N_i(2leq ileq Q)Ni(2iQ) is defined as the xor of the previous N_{i-1}Ni1 and the square of the previous answer A_{i-1}Ai1. For example, if the first query N_1N1 is 22, the answer A_1A1 is 33, then the second query N_2N2 is 2 xor ( 3*3)=112 xor (33)=11.

    Finally, you don't need to output all the answers for every query, you just need to output the xor of each query's answer A_1 xor A_2 ... xor A_QA1 xor A2...xor AQ.

    Input

    The input contains two integers, Q, NQ,N1 leq Q leq 10^7,0 leq N leq 10^{18}1  Q107,0  N1018QQ representing the number of queries and NN representing the first query.

    Output:

    An integer representing the final answer.

    样例输入

    17 473844410

    样例输出

    903193081
    所遇皆星河
  • 相关阅读:
    自习任我行第二阶段个人总结9
    自习任我行第二阶段个人总结8
    自习任我行第二阶段个人总结7
    自习任我行第二阶段个人总结6
    自习任我行第二阶段个人总结5
    自习任我行 第二阶段每日个人总结4
    自习任我行 第二阶段每日个人总结3
    自习任我行 第二阶段每日个人总结2
    自习任我行 第二阶段每日个人总结1
    结课总结
  • 原文地址:https://www.cnblogs.com/Shallow-dream/p/11488867.html
Copyright © 2011-2022 走看看