POJ3070 斐波那契数列 矩阵快速幂

题目链接：http://poj.org/problem?id=3070

题意就是让你求斐波那契数列，不过n非常大，只能用logn的矩阵快速幂来做了

刚学完矩阵快速幂刷的水题，POJ不能用万能头文件是真的烦

#include<iostream>
#include<string.h>
#include<cmath>
#include<cstdio>
using namespace std;
typedef long long ll;
const int inf=<<;
const int maxn=1e5+;
const double pi=acos(-);
const int mod=;
struct matrix{
    ll a[][]; //begin with 1
    int r,c;
    matrix(int n,int m):r(n),c(m){memset(a,,sizeof(a));}
    ll* operator[](int x){return a[x];}
    friend matrix operator*(matrix A,matrix B)
    {
        matrix C(A.r,B.c);
        for(int i=;i<=A.r;i++)
            for(int j=;j<=B.c;j++)
                for(int k=;k<=A.c;k++){
                    C[i][j]+=(A[i][k]*B[k][j])%mod;
                    C[i][j]+=mod;
                    C[i][j]%=mod;
                }
        return C;
    }
};

matrix qpow(matrix A,ll m)//方阵A的m次幂
{
    matrix ans(A.r,A.c);
    for(int i=;i<=A.r;i++) ans.a[i][i]=; //单位矩阵
    while(m)
    {
        if(m&)ans=ans*A;
        A=A*A;
        m>>=;
    }
    return ans;
}
/*inline void read(ll &x){
    char ch=x=0;
    while(!isdigit(ch))
        ch=getchar();
    while(isdigit(ch))
        x=x*10+ch-'0',ch=getchar();
}*/
int main(){
    ll n;
    while(~scanf("%lld",&n)){
        if(n==-)break;
        if(n==){
            cout<<<<endl;continue;
        }
        matrix a(,);
        a[][]=,a[][]=,a[][]=,a[][]=;
        matrix x1(,);
        x1[][]=,x1[][]=;
        matrix ans=qpow(a,n-)*x1;//0cout<<233<<endl;
        printf("%d\n",ans[][]%);
    }
    return ;
}