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）

  1 #include<bits/stdc++.h>
  2 using namespace std;
  3 //const int maxn=1000005;
  4 using namespace std;
  5 typedef  long  long  ll;
  6 const int N = 2;//矩阵大小
  7 //ll k;
  8 const long long mod=(long long )998244353;
  9 struct Mat
 10 {
 11     ll mat[N][N];
 12     Mat operator*(const Mat a)const
 13     {
 14         Mat b; memset(b.mat, 0, sizeof(b.mat));
 15         for (int i = 0; i < N; i++)
 16             for (int j = 0; j < N; j++)
 17                 for (int k = 0; k < N; k++)
 18                     b.mat[i][j] = (b.mat[i][j] + (mat[i][k]) *(a.mat[k][j])) % mod;
 19         return b;
 20     }
 21 };
 22 
 23 ll phi(ll x)//求欧拉
 24 {
 25     ll res=x;
 26     for(ll i=2;i*i<=x;i++)
 27     {
 28         if(x%i==0)
 29         {
 30             res=res/i*(i-1);
 31             while(x%i==0) x/=i;
 32         }
 33     }
 34     if(x>1) res=res/x*(x-1);
 35     return res;
 36 }
 37 Mat Pow(Mat m, ll k)
 38 {
 39     //if(k==1) return 1;
 40     Mat ans;
 41     memset(ans.mat, 0, sizeof(ans.mat));
 42     for (int i = 0; i < N; i++)
 43         ans.mat[i][i] = 1;
 44     while (k)
 45     {
 46         if (k & 1) ans = ans*m;
 47         k >>= 1;
 48         m = m*m;
 49     }
 50     return ans;
 51 }
 52 
 53 ll que[4*2000000];
 54 int head=0,tail=1 ;
 55 int main() {
 56 
 57     ll phi_mod=phi(mod);
 58     Mat m;
 59     int q;
 60     ll n,ans;
 61     scanf("%d%lld",&q,&n);
 62     //printf("\n%d %lld\n",q,n);
 63     Mat f;
 64     m.mat[0][0]=3;m.mat[0][1]=2;
 65     m.mat[1][0]=1;m.mat[1][1]=0;
 66     f.mat[0][0]=1;//x1
 67     f.mat[1][0]=0;//x0
 68     f = Pow(m, (n-1)%phi_mod)*f;
 69     ans=f.mat[0][0];
 70     //printf("%lld %lld\n",n,ans);
 71     que[head]=ans;
 72 
 73 
 74 
 75 
 76     //printf("%lld",f.mat[0][0]);
 77 /*
 78  * 10000000 1000000000000000000
 79  * */
 80     for (register int i = 2; i <= q; i++) {
 81         //init(m);
 82         //memset(f.mat, 0, sizeof(f.mat));
 83         n = n ^ (f.mat[0][0] * f.mat[0][0]);
 84 
 85         if (n == 0) {
 86             f.mat[0][0] = 0;
 87         } else if (n == 1) {
 88             f.mat[0][0] = 1;
 89         } else {
 90             m.mat[0][0] = 3;
 91             m.mat[0][1] = 2;
 92             m.mat[1][0] = 1;
 93             m.mat[1][1] = 0;
 94             f.mat[0][0] = 1;//x1
 95             f.mat[1][0] = 0;//x0
 96 
 97             f = Pow(m, (n - 1) % phi_mod) * f;
 98 
 99         }
100         ans = ans ^ f.mat[0][0];
101         //printf("%lld %lld\n", n, ans);
102 
103 
104         que[tail++]=ans;
105         if(tail-head>4)
106         {
107             head++;
108             if(que[head]==que[head+2]&&que[head+1]==que[head+1+2])
109             {
110                 int tmp=q-i;
111                 if(tmp%2)
112                 {
113                     ans=que[head];
114                 }
115                 else
116                 {
117                     ans=que[head+1];
118                 }
119                 break;
120             }
121         }
122 
123 
124     }
125     printf("%lld\n",ans);
126     return 0;
127 }
128 
129 /*
130  * 858251072
131  *
132  * 245284867829898842 447003402
133     485245887812443738 1008229130
134  *
135  *
136  *
137  *
138  * */