H. The Nth Item（The 2019 Asia Nanchang First Round Online Programming Contest）

题意：https://nanti.jisuanke.com/t/41355

给出N1，计算公式：A=F(N）Ni=Ni-1 ^ (A*A)，F为类斐波那契需要矩阵快速幂的递推式。

求第k个N。

思路：

发现从大约1e5个数开始N交替出现，到一定位置%2即可。（or正解：https://blog.csdn.net/qq_41848675/article/details/100667808   or

https://blog.csdn.net/jk_chen_acmer/article/details/100635672   or   map记忆化）

 #include<bits/stdc++.h>
 using namespace std;
 //const int maxn=1000005;
 using namespace std;
 typedef  long  long  ll;
 const int N = ;//矩阵大小
 //ll k;
 const long long mod=(long long );
 struct Mat
 {
     ll mat[N][N];
     Mat operator*(const Mat a)const
     {
         Mat b; memset(b.mat, , sizeof(b.mat));
         for (int i = ; i < N; i++)
             for (int j = ; j < N; j++)
                 for (int k = ; k < N; k++)
                     b.mat[i][j] = (b.mat[i][j] + (mat[i][k]) *(a.mat[k][j])) % mod;
         return b;
     }
 };

 ll phi(ll x)//求欧拉
 {
     ll res=x;
     for(ll i=;i*i<=x;i++)
     {
         if(x%i==)
         {
             res=res/i*(i-);
             while(x%i==) x/=i;
         }
     }
     if(x>) res=res/x*(x-);
     return res;
 }
 Mat Pow(Mat m, ll k)
 {
     //if(k==1) return 1;
     Mat ans;
     memset(ans.mat, , sizeof(ans.mat));
     for (int i = ; i < N; i++)
         ans.mat[i][i] = ;
     while (k)
     {
         if (k & ) ans = ans*m;
         k >>= ;
         m = m*m;
     }
     return ans;
 }

 ll que[*];
 int head=,tail= ;
 int main() {

     ll phi_mod=phi(mod);
     Mat m;
     int q;
     ll n,ans;
     scanf("%d%lld",&q,&n);
     //printf("\n%d %lld\n",q,n);
     Mat f;
     m.mat[][]=;m.mat[][]=;
     m.mat[][]=;m.mat[][]=;
     f.mat[][]=;//x1
     f.mat[][]=;//x0
     f = Pow(m, (n-)%phi_mod)*f;
     ans=f.mat[][];
     //printf("%lld %lld\n",n,ans);
     que[head]=ans;

     //printf("%lld",f.mat[0][0]);
 /*
  * 10000000 1000000000000000000
  * */
     for (register int i = ; i <= q; i++) {
         //init(m);
         //memset(f.mat, 0, sizeof(f.mat));
         n = n ^ (f.mat[][] * f.mat[][]);

         if (n == ) {
             f.mat[][] = ;
         } else if (n == ) {
             f.mat[][] = ;
         } else {
             m.mat[][] = ;
             m.mat[][] = ;
             m.mat[][] = ;
             m.mat[][] = ;
             f.mat[][] = ;//x1
             f.mat[][] = ;//x0

             f = Pow(m, (n - ) % phi_mod) * f;

         }
         ans = ans ^ f.mat[][];
         //printf("%lld %lld\n", n, ans);

         que[tail++]=ans;
         if(tail-head>)
         {
             head++;
             if(que[head]==que[head+]&&que[head+]==que[head++])
             {
                 int tmp=q-i;
                 if(tmp%)
                 {
                     ans=que[head];
                 }
                 else
                 {
                     ans=que[head+];
                 }
                 break;
             }
         }

     }
     printf("%lld\n",ans);
     return ;
 }

 /*
  * 858251072
  *
  * 245284867829898842 447003402
     485245887812443738 1008229130
  *
  *
  *
  *
  * */