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\n",dns);
}

 

For a series $F$:

$\begin{array}{c}F(0)=0,F(1)=1\\ F(n)=3\ast F(n-1)+2\ast F(n-2),(n\ge 2)\end{array}$

 

We have some queries. For each query $N$, the answer $A$ is the value $F(N)$ modulo $998244353$.

Moreover, the input data is given in the form of encryption, only the number of queries $Q$ and the first query $N^1$are given. For the others, the query $N^i(2\le i\le Q)$ is defined as the xor of the previous $N^{i-1}$ and the square of the previous answer $A^{i-1}$. For example, if the first query $N^1$ is $2$, the answer $A^1$ is $3$, then the second query $N^2$ is $2xor(3\ast 3)=11$.

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 $A^{1}xor{A}^2...xor{A}^Q$.

Input

The input contains two integers, $Q,N$, $1\le Q\le 10^7,0\le N\le 10^{18}$. $Q$ representing the number of queries and $N$ representing the first query.

Output:

An integer representing the final answer.

样例输入复制

17 473844410

样例输出复制

903193081