[Matrix Multiplication] [SSL 1529] Fibonacci Sequence Ⅱ

[Matrix Multiplication] [SSL 1529] Fibonacci Sequence Ⅱ

topic

Shaped like 1,123,581,321,345,589,144 ... number of columns, the number of columns required Pei Bo Laqi of $n$ items.

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enter

$n$ (1〈 $n$ 〈2 ^ 31）

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Output

A number of series of Pei Bola Qi $n$ items$mod$ 10000;

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Sample

input
123456789

output
4514

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Problem-solving ideas

Matrix multiplication
Two matrices $A$,$B$ multiplied by
tm tmThe length of the product of$t$$m$ is$A$ 's length and width areBBWideAA of$B$
The sum of the product of the number in the first row of$A$ multiplied by the number in the first column of B is the number in the first row and first column of the product
.

The writing method of this question is
given that $f[n]=f[n-1]+f[n-2]$
If you require$f[n+1]$
即$f[n+1]=f\left[n\right]+f[n-1]$
Expand$f[n+1]=f[n-1]+f[n-2]+f[n-1]$
An answer matrix can be constructed to store the current$f[n-1]$ and$f[n]\; The$
current f[n] is the next$f[n-1]$
and the next$f\left[n\right]$ is the current f[n] and$f[n-1]$ The sum
can be constructed as a matrix. In the

title,$n$ is 2^31, it is
conceivable to use fast exponentiation to solve

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Code

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const int mo=10000;
long long n;
struct hhx{
    
    
	long long n,m,h[5][5];
}a,b,x;
hhx operator * (hhx l,hhx y)  //重新定义乘号
{
    
    
	x.n=l.n,x.m=y.m;
	memset(x.h,0,sizeof(x.h));
	for (int k=1;k<=l.m;k++)
	    for (int i=1;i<=x.n;i++)
		    for (int j=1;j<=x.m;j++) 
		        x.h[i][j]=(x.h[i][j]+l.h[i][k]*y.h[k][j]%mo)%mo;
	return x;
}
void power(long long n)  //快速幂
{
    
     
	 if (n & 1) a=a*b;
	 if (n==1) return;
	 b=b*b; 
	 power(n/2);
}
int main()
{
    
    
	a.n=1,a.m=2;
	a.h[1][1]=a.h[1][2]=1;
	b.n=2,b.m=2;
	b.h[1][2]=b.h[2][1]=b.h[2][2]=1;  //赋值
	scanf("%lld",&n);
	power(n-1);
	printf("%lld",a.h[1][1]);
	return 0;
}