链接：

 

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11236		Accepted: 7991

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

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:

.

 

代码：

#include
       
        #include
        
         #define MOD 10000struct node{    int m[2][2];}a, b;node cheng(node x, node y){    int i, j, k;    node c;    for(i=0; i<2; i++)    for(j=0; j<2; j++)    {        c.m[i][j] = 0;        for(k=0; k<2; k++)            c.m[i][j] = (c.m[i][j] + x.m[i][k]*y.m[k][j])%MOD;    }    return c;}int Fast_MOD(int n) {    a.m[0][0] = a.m[0][1] = a.m[1][0] = 1;    a.m[1][1] = 0;    b.m[0][0] = b.m[1][1] = 1;  /// b 初始化为单位矩阵    b.m[0][1] = b.m[1][0] = 0;    while(n)    {        if(n&1) /// n是奇数            b = cheng(b, a);        a = cheng(a, a);        n >>= 1;    }    return b.m[0][1];}int main(){    int n;    while(scanf("%d", &n), n!=-1)    {        printf("%d\n", Fast_MOD(n));    }    return 0;}