POJ 3070 Fibonacci 矩阵乘法

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title

POJ 3070
OJ 1351
LUOGU 1962两题 $mod$略有不同
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

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:

Source

Stanford Local 2006

analysis

YYF.

code

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod=1e4;
template<typename T>inline void read(T &x)
{
	x=0;
	T f=1, ch=getchar();
	while (!isdigit(ch) && ch^'-') ch=getchar();
	if (ch=='-') f=-1, ch=getchar();
	while (isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48), ch=getchar();
	x*=f;
}
struct matrix
{
	ll num[2][2];
	inline void build()
	{
		for (int i=0; i<2; ++i)
			for (int j=0; j<2; ++j)
				num[i][j]=1;
		num[1][1]=0;
	}
};
inline matrix mul(matrix a,matrix b)//矩阵乘法
{
	matrix res;
    memset(res.num,0,sizeof(res));
    for(int i=0; i<2; ++i)
        for(int j=0; j<2; ++j)
            for(int k=0; k<2; ++k)
                (res.num[i][j]+=a.num[i][k]*b.num[k][j])%=mod;
    return res;
}
inline void Quick_power(ll p)//矩阵快速幂
{
	matrix a,res;
	a.build();
    memset(res.num,0,sizeof(res));
    for (int i=0; i<2; ++i)
		res.num[i][i]=1;
    while (p)
    {
        if (p&1)
			res=mul(res,a);
        a=mul(a,a);
        p>>=1;
    }
    printf("%lld\n",res.num[0][1]);
}
int main()
{
    while (1)
    {
    	ll n;read(n);
		if (n==-1) break;
		Quick_power(n);//计算基础矩阵
	}
    return 0;
}