POJ 3070(Fibonacci-矩阵幂)

Default Fibonacci Description  Fibonacci integer sequence指 F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 2 for n ≥ 2. eg: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, … Fibonacci 的矩阵乘法求法如下 . Given an integer n, 求 Fn mod 10000. Input 多组数据,每行一个 n (where 0 ≤ n ≤ 1,000,000,000). 数据以 −1 结尾. Output 对每组数据打印一行 Fn mod 10000). Sample Input 099999999991000000000-1 Sample Output 0346266875 Hint As a reminder, matrix multiplication is associative, and the product of two 2 × 2 matrices is given by . Also, note that raising any 2 × 2 matrix to the 0th power gives the identity matrix: . Source ​​Stanford Local 2006​​

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6733		Accepted: 4765

正宗矩阵幂练手题。

做了这题就差不多理解矩阵乘法了。

补充：递归一般能用矩阵乘法，如f[i]=g(f[i-1],f[i-2])...

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<functional>
using namespace std;
#define MAXN (1000000000)
#define F (10000)
struct M
{
  int a[3][3];
  M(int i){a[1][1]=a[1][2]=a[2][1]=1;a[2][2]=0; }
  M(){memset(a,0,sizeof(a));  }
  friend M operator*(M a,M b)
  {
    M c;
    for (int i=1;i<=2;i++)
      for (int j=1;j<=2;j++)
        for (int k=1;k<=2;k++)
        {
          c.a[i][j]=(c.a[i][j]+a.a[i][k]*b.a[k][j])%F;
        }
    return c;
  }
  friend M pow(M a,int b)
  {
    if (b==1)
    {
      M c(1);
      return c;
    }
    else 
    {
      M c=pow(a,b/2);
      c=c*c;
      if (b%2) return c*a;
      return c;
    }
  }
};
int n;
int main()
{
  while (cin>>n&&n!=-1)
  {
    if (n==0) printf("0\n");
    else
    {
      M a=pow(M(1),n);
      printf("%d\n",a.a[1][2]);
    }
  }
}