斐波那契数列

http://blog.youkuaiyun.com/lvshubao1314/article/details/38013897

一：递推

二：矩阵

//斐波那契数列第N项后MOD位，矩阵快速幂求法； 
#include<iostream> 
#include<cstdio>
#include<string.h>
#define LL long long
using namespace std;
const int MOD=10000;
struct Matrix
{
	LL m[3][3];
};
Matrix A={0,0,0,
		  0,1,1,
		  0,1,0,};
Matrix E={0,0,0,
		  0,1,0,
		  0,0,1,};
Matrix multi(Matrix a,Matrix b)
{
	Matrix ans;
	for(int i=1;i<=2;i++)
	{
		for(int j=1;j<=2;j++)
		{
			ans.m[i][j]=0;
			for(int k=1;k<=2;k++)
				ans.m[i][j]+=a.m[i][k]*b.m[k][j]%MOD;
			ans.m[i][j]%=MOD;
		}
	}
	return ans;
}
Matrix power(Matrix a,int p)
{
	Matrix ans=E,B=a;
	while(p)
	{
		if(p%2==1)
			ans=multi(ans,B);
		p=p/2;
		B=multi(B,B);
	} 
	return ans;
}
int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		Matrix ans=power(A,n-1);
		if(n==0)
			printf("0\n");
		else
			printf("%d\n",ans.m[1][1]);
	}
	return 0;
}

三：利用数学对数计算求前几位数字

#include<iostream> 
#include<cstdio>
#include<cmath>
using namespace std;
int main()
{
	int f[25];
	f[0]=0,f[1]=1;
	for(int i=2;i<=20;i++)
	{
		f[i]=f[i-1]+f[i-2];
	}
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		if(n<=20)
			printf("%d\n",f[n]);
		else
		{
			double f=(1.0+sqrt(5.0))/2.0;
			double p=-0.5*log10(5.0)+n*1.0*log(f)/log(10.0);
			p=p-floor(p);
			double ans=pow(10.0,p);
			while(ans<1000)
				ans=ans*10;
			printf("%d\n",(int)ans);
			
		}
	}
	return 0;
}