ACdream 1060 递推数

Problem Description

已知A(0) = 0 , A(1) = 1 , A(n) = 3 * A(n-1) + A(n-2) (n ≥ 2)
求 A(A(A(A(N)))) Mod (1e9 + 7)

Input

第一行一个整数 T (T ≤ 10000) 代表数据组数
每组数据占一行，一个整数 n (1 ≤ n ≤ 1e12)

Output

对于每组测试数据输出一个整数。

Sample Input

4
1
23574
27870
913

Sample Output

1
0
0
1

code

/** this code is made by linglian* Problem: 1060* Verdict: Accepted* Submission Date: 2017-05-05 20:38:43* Time: 300MS* Memory: 1664KB*/#include <stdio.h>  #include <string.h>  #include <iostream>  #include <algorithm>  #include <math.h>  using namespace std;  const long long mod1=1e9+7;//循环节  const long long mod2=222222224;  const long long  mod3=183120;  const long long  mod4=240;  struct Matrix  {      long long mapp[2][2];  };  Matrix p= {0,1,1,0}; //左边矩阵  Matrix p1= {0,1,1,3};//公式矩阵  Matrix unin= {1,0,0,1};//单位矩阵  Matrix A(Matrix a,Matrix b,long long mod)  {      Matrix c;      for(int i=0; i<2; i++)          for(int j=0; j<2; j++)          {              c.mapp[i][j]=0;              for(int k=0; k<2; k++)                  c.mapp[i][j]+=(a.mapp[i][k]*b.mapp[k][j])%mod;                  c.mapp[i][j]%=mod;          }      return c;  }  Matrix A(long long n,long long mod) {      Matrix m=p1,b=unin;      while(n)      {          if(n&1) b=A(b,m,mod);           n>>=1;           m=A(m,m,mod);      }      return A(p,b,mod);  }  long long n;  int main() {      int T;      scanf("%d",&T);      while(T--)      {          scanf("%lld",&n);          Matrix ans;          ans=A(n,mod4);        ans=A(ans.mapp[0][0],mod3);          ans=A(ans.mapp[0][0],mod2);          ans=A(ans.mapp[0][0],mod1);          printf("%lld\n",ans.mapp[0][0]);      }      return 0;  }