poj 3070 Fibonacci

 

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2134		Accepted: 1471

Description

In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, and F_n = F_{n − 1} + F_{n − 2} for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …

An alternative formula for the Fibonacci sequence is

.

Given an integer n, your goal is to compute the last 4 digits of F_n.

Input

The input test file will contain multiple test cases. Each test case consists of a single line containing n (where 0 ≤ n ≤ 1,000,000,000). The end-of-file is denoted by a single line containing the number −1.

Output

For each test case, print the last four digits of F_n. If the last four digits of F_n are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print F_n mod 10000).

Sample Input

0

9

999999999

1000000000

-1

Sample Output

0

34

626

6875

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

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/* 矩阵的方法。 
{{F(n+1),F(n)},
{F(n),F(n-1)}}={{1,1}      ^n
                        {1,0}}

令A(n)表示一个矩阵序列

A(n)={{F(n),F(n-1)},
         {F(n-1),F(n-2)}那么A(2)={{1,1},{1,0}}，由那个表达式得到：A(n)=A(2)^(n-1)，A(2)是己知的2*2矩阵，现在的问题就是如何求

A(2)^n因为方阵的乘法有结合律，所以A(2)^n=A(2)^(n/2)*A(2)^(n/2)，不妨设n是偶数

所以求A(n)就可以化成求A(n/2)并作一次乘法，所以递归方程是：T(n)=T(n/2)+O(1)
*/

// 5144566 11410 3070 Accepted 204K 0MS C++ 1049B 2009-05-13 10:13:17
// 用矩阵做Fibonacci数 
#include  < iostream >
#define  MAX 2
using   namespace  std;
typedef  struct  node
{
     int  matrix[MAX][MAX];
}Matrix;
Matrix unit,init,result;
int  n;
void  Init()   // 初始化
{
     int  i,j;
     for (i = 0 ;i < 2 ;i ++ )
         for (j = 0 ;j < 2 ;j ++ )
        {
            init.matrix[i][j] = 1 ;
            unit.matrix[i][j] = (i == j);
        }
    init.matrix[ 1 ][ 1 ] = 0 ;
}
Matrix Mul(Matrix a,Matrix b)  // 矩阵乘法
{
    Matrix c;
     int  i,j,k;
     for (i = 0 ;i < 2 ;i ++ )
         for (j = 0 ;j < 2 ;j ++ )
        {
            c.matrix[i][j] = 0 ;
             for (k = 0 ;k < 2 ;k ++ )
                c.matrix[i][j] += a.matrix[i][k] * b.matrix[k][j];
             if (c.matrix[i][j] >= 10000 )   // 取余
                c.matrix[i][j] %= 10000 ;
        }
     return  c;
}
Matrix Cal( int  exp)   // 求幂
{
    Matrix p,q;
    p = unit;
    q = init;
     while (exp != 1 )
    {
         if (exp & 1 )
        {
            exp -- ;
            p = Mul(p,q);
        }
         else
        {
            exp >>= 1 ;
            q = Mul(q,q);
        }
    }
    p = Mul(p,q);
     return  p;
}
int  main()
{
     while (scanf( " %d " , & n) != EOF && n !=- 1 )
    {
         if (n == 0 )
        {
            printf( " 0\n " );
             continue ;
        }
         if (n == 2 || n == 1 )
        {
            printf( " 1\n " );
             continue ;
        }
        Init();
        result = Cal(n - 1 );   // 求n-1次幂就好
        printf( " %d\n " ,result.matrix[ 0 ][ 0 ]);
    }
     return   0 ;
}