[poj3070][Fibonacci] (矩阵快速幂)

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:

.

Solution

矩阵快速幂

#include<stdio.h>
#include<string.h>
#define md 10000
long long n;
struct matrix{
    long long x[2][2];
    matrix operator*(matrix b){
        matrix tmp;
        for(int i=0;i<2;i++)
            for(int j=0;j<2;j++){
                tmp.x[i][j]=0;
                for(int k=0;k<2;k++)
                    tmp.x[i][j]=(tmp.x[i][j]+x[i][k]*b.x[k][j])%md;
            }
        return tmp;
    }
}root,counter;
void matrix_qpow(long long p){
    for(;p;counter=counter*counter,p>>=1)
        if(p&1)
            root=root*counter;
}
int main(){
    scanf("%I64d",&n);
    while(n!=-1){
        counter.x[0][0]=counter.x[0][1]=counter.x[1][0]=1;
        counter.x[1][1]=0;
        root.x[0][0]=root.x[1][1]=1;
        root.x[0][1]=root.x[1][0]=0;
        matrix_qpow(n);
        printf("%I64d\n",root.x[0][1]);
        scanf("%I64d",&n);
    }
    return 0;
}