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;
}