题意
求斐波那契数列第n项,n<=1e18
solution
1.显然O(n)递推肯定不行
2.所以我们考虑用矩阵快速幂加速递推
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<cmath>
#include<cstdlib>
#include<ctime>
using namespace std;
typedef long long ll;
const ll inf=0x3f3f3f3f;
const ll p=100000007;
inline ll read(){
char ch=' ';ll f=1;ll x=0;
while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}
while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
return x*f;
}
struct Matrix
{
ll x,y;
ll m[3][3];
}a,b,c;
Matrix operator *(Matrix a,Matrix b)
{
ll n,m,q;
n=a.x;m=a.y;
q=b.y;
Matrix tmp;
tmp.x=n;tmp.y=q;
memset(tmp.m,0,sizeof(tmp.m));
for(ll i=1;i<=n;i++)
{
for(ll j=1;j<=m;j++)
{
for(ll k=1;k<=q;k++)
{
tmp.m[i][j]+=a.m[i][k]*b.m[k][j];
tmp.m[i][j]%=p;
}
}
}
return tmp;
}
Matrix mul(Matrix x,ll y)
{
Matrix ret;
ret.m[1][1]=1;
ret.m[1][2]=0;
ret.m[2][1]=0;
ret.m[2][2]=1;
ret.x=2;ret.y=2;
while(y)
{
if(y&1)
{
ret=ret*x;
}
x=x*x;
y=y>>1;
}
return ret;
}
int main()
{
ll n=read();
a.m[1][1]=1;
a.m[1][2]=1;
a.m[2][1]=1;
a.m[2][2]=0;
a.x=2;a.y=2;
b.m[1][1]=1;
b.m[1][2]=1;
b.x=1;b.y=2;
c=mul(a,n);
b=b*c;
cout<<b.m[1][1]<<endl;
return 0;
}