P1962 斐波那契数列 题解

Link

P1962 斐波那契数列

Solve

(O(n))的递推显然是要挂的，考虑矩阵乘法

推出式子就好了，实在模板

注意要从(n-2)开始乘

(n<2)的特判一下

Code

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL TT=1e9+7;
LL N;
struct AS{
	LL a[3][3];
	AS(){memset(a,0,sizeof a);}
	void clear(){memset(a,0,sizeof a);a[2][1]=a[1][1]=a[1][2]=1;}
	AS operator *(const AS B){
		AS C;
		for(int i=1;i<3;i++)
		for(int j=1;j<3;j++)
		for(int k=1;k<3;k++)
			C.a[i][j]=(C.a[i][j]+a[i][k]*B.a[k][j]%TT)%TT;
		return C;
	}
};
int main(){
	freopen("P1962.in","r",stdin);
	freopen("P1962.out","w",stdout);
	scanf("%lld
",&N);
	if(N==1){printf("%d
",1);return 0;}
	if(N==0){printf("%d
",0);return 0;}
	AS s,w;s.a[1][1]=s.a[1][2]=1;w.clear();
	N-=2;
	while(N){
		if(N&1)s=s*w;
		w=w*w;N>>=1;
	}
	printf("%lld
",s.a[1][1]%TT);
	return 0;
}