Fibonacci POJ

In the Fibonacci integer sequence, F₀ = 0,F₁ = 1, andF_n =F_{n − 1} +F_{n − 2} forn ≥ 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 ofF_n are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., printF_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:

.

做这个题真的是开眼了，以前不知道有矩阵快速幂QAQ，看来矩阵快速幂是可以优化递推关系的

操作过程如同对数的快速幂，我自己推了一遍矩阵表达式，推出来的不是给出的那个，而是如下这个

（图是抄的）

这样的话就要>=1,当n是0的时候直接输出0就可以了。还有，用结构体里存放矩阵的话非常清晰方便，比不用来的要清晰简单得多。

#include<stdio.h>#include<string.h>using namespace std;struct aa{    int m[3][3];};aa mx(aa a,aa b){    aa res;    memset(res.m,0,sizeof(res.m));    for(int i=1; i<=2; i++)        for(int j=1; j<=2; j++)            for(int k=1; k<=2; k++)            {                res.m[i][j]+=(a.m[i][k]*b.m[k][j]);                res.ma[i][j]%=10000;            }    return res;}int fast_mod(aa x,long long y){    aa ans;    ans.m[1][1]=ans.m[2][2]=1;    ans.m[1][2]=ans.m[2][1]=0;    while(y)    {        if(y&1)        {            ans=mx(ans,x);        }        y=y>>1;        x=mx(x,x);    }    return ans.m[1][1];}int main(){    long long n;    while(scanf("%lld",&n)!=EOF)    {        if(n<0)            break;        aa x;        x.m[1][1]=x.m[1][2]=x.m[2][1]=1;        x.m[2][2]=0;        if(n==0)            printf("0\n");        else        printf("%d\n",fast_mod(x,n-1));    }}