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; }