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.

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.

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).

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:

.

#include <iostream>#include <cstdio>#include <time.h>using namespace std;#define MOD 10000#define N 2void mul(int a[N][N],int b[N][N]){    int i,j,k;    int c[N][N]={0};    for(i=0;i<N;i++)    {        for(j=0;j<N;j++)        {            for(k=0;k<N;k++)            {                c[i][j]=(c[i][j]+a[i][k]*b[k][j])%MOD;            }        }    }    for(i=0;i<N;i++)    {        for(j=0;j<N;j++)        {            a[i][j]=c[i][j];        }    }}int main(){    int n;    while(scanf("%d",&n),n+1)    {        int a[N][N]={{0},{1}},b[N][N]={{1,1},{1,0}};     //首项和递推矩阵        while(n)                                         //二分快速幂        {            if(n&1) mul(a,b);            mul(b,b);            n>>=1;        }        printf("%d\n",a[1][1]);    }    return 0;}