更新时间:2022-08-12 16:43:39
思路: 矩阵快速幂
分析:
1 最简单的矩阵快速幂
代码:
/************************************************ * By: chenguolin * * Date: 2013-08-28 * * Address: http://blog.csdn.net/chenguolinblog * ************************************************/ #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; const int MOD = 10000; const int N = 2; int n; struct Matrix{ int mat[N][N]; Matrix operator*(const Matrix& m)const{ Matrix tmp; for(int i = 0 ; i < N ; i++){ for(int j = 0 ; j < N ; j++){ tmp.mat[i][j] = 0; for(int k = 0 ; k < N ; k++) tmp.mat[i][j] += mat[i][k]*m.mat[k][j]%MOD; tmp.mat[i][j] %= MOD; } } return tmp; } }; int Pow(Matrix m){ if(n <= 1) return n; Matrix ans; memset(ans.mat , 0 , sizeof(ans.mat)); for(int i = 0 ; i < N ; i++) ans.mat[i][i] = 1; n--; while(n){ if(n%2) ans = ans*m; n /= 2; m = m*m; } return ans.mat[0][0]%MOD; } int main(){ Matrix m; m.mat[0][0] = m.mat[0][1] = 1; m.mat[1][0] = 1 ; m.mat[1][1] = 0; while(scanf("%d" , &n) && n != -1) printf("%d\n" , Pow(m)); return 0; }