poj 3070 Fibonacci (矩阵快速幂乘/模板)

Posted 2020-09-07 simpleknight

题意：给你一个n,输出Fibonacci (n)%10000的结果

思路：裸矩阵快速幂乘，直接套模板

代码：

#include <cstdio>
#include <cstring>
#include <iostream>

using namespace std;

typedef long long ll;
const int N=2,M=2,P=2;
const int MOD=10000;
struct  Matrix
{
    ll m[N][N];
};

Matrix  A={1,1,
          1,0};

Matrix  I={1,0,
          0,1};

Matrix multi(Matrix a,Matrix b)
{
    Matrix ans;
    for(int i=0;i<N;i++)
    {
       for(int j=0;j<M;j++)
       {
          ans.m[i][j]=0;
          for(int k=0;k<P;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 k)
{
    Matrix ans=I,p=a;
    while(k)
    {
      if(k&1)
      {
        ans=multi(ans,p);
      }
      k>>=1;
      p=multi(p,p);
    }
    return ans;
}

int main(int argc, char const *argv[])
{
    int n;
    while(scanf("%d",&n)!=-1)
    {
       if(n==-1) break;
       if(n==0)
       {
           cout<<"0"<<endl;
           continue;
       }
       Matrix ans=power(A,n-1);
       printf("%lld\n",ans.m[0][0] );
    }
    return 0;
}