Fibonacci（矩阵快速幂入门）

Description

In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 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 Fn.

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 Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn mod 10000).

Sample Input

0
9
999999999
1000000000
-1
Sample Output

0
34
626
6875

矩阵快速幂入门，照着给的公式用一下就行
本题让没接触矩阵运算的我对矩阵乘法，单位矩阵有了一些认识

AC代码

#include"iostream"
#include"string.h"
#include"algorithm"
using namespace std;
typedef long long ll;
struct MUL{
	ll mul[10][10];
}start;

MUL ksm(MUL x,MUL y){
	MUL z;
	memset(z.mul,0,sizeof(z.mul));
	for(int i=1;i<=2;i++){
		for(int j=1;j<=2;j++){
			for(int k=1;k<=2;k++){
				z.mul[i][j] += x.mul[i][k] * y.mul[k][j]%10000;//取后4位
				z.mul[i][j] %= 10000;
			}
		}
	}
	return z;
}

ll jzksm(MUL x,ll n){
	MUL res;
	memset(res.mul,0,sizeof(res.mul));
	for(int i=1;i<=2;i++){//单位矩阵对角线赋值1，其余全为0，单位矩阵相当于数值中的1 
		res.mul[i][i]=1;
	}
	while(n){//快速幂，不知道的小伙伴请百度一下
		if(n&1){
			res=ksm(x,res);
		}
		x=ksm(x,x);
		n>>=1;
	}
	return res.mul[1][2]%10000;
}
int main()
{
	
	ll w;
	while(cin>>w){
		if(w==-1){
			break;
		}
		start.mul[1][1]=start.mul[1][2]=start.mul[2][1]=1;
		start.mul[2][2]=0;
		cout<