POJ 3070 + 51Nod 1242 大斐波那契数取余

Posted 2020-09-28 kimsimple

 

 POJ 3070

 

 

#include "iostream"
#include "cstdio"
using namespace std;

class matrix
{
public:
    int a[2][2];
    matrix()
    {
        a[0][0]=a[1][0]=a[0][1]=1;
        a[1][1]=0;
    }
};

matrix multi(matrix a,matrix b)
{
    matrix temp;
    int i,j,k;
    for(i=0;i<2;i++)
        for(j=0;j<2;j++)
        {
            temp.a[i][j]=0;
            for(k=0;k<2;k++)
                temp.a[i][j]+=(a.a[i][k]*b.a[k][j]);
            temp.a[i][j]%=10000;
        }
        return temp;
}


matrix power(int n)
{
     matrix temp,s;
        temp.a[0][1]=temp.a[1][0]=0;
        temp.a[0][0]=temp.a[1][1]=1;
        while(n!=0)
        {
          if(n%2!=0)
            temp=multi(temp,s);
        s=multi(s,s);
        n=n/2;
        }
        return temp;
}

int main()
{
    int n;
    while(~scanf("%d",&n)&&(n!=-1))
    {

        matrix t=power(n);
        cout<<t.a[1][0]<<endl;
    }
    return 0;
}

 

 

 51Nod 1242 大斐波那契数取余

#include "iostream"
#include "cstdio"
using namespace std;
#define MOD 1000000009
#define LL long long
class matrix
{
public:
    LL a[2][2];
    matrix()
    {
        a[0][0]=a[1][0]=a[0][1]=1;
        a[1][1]=0;
    }
};

matrix multi(matrix a,matrix b)
{
    matrix temp;
    LL i,j,k;
    for(i=0;i<2;i++)
        for(j=0;j<2;j++)
        {
            temp.a[i][j]=0;
            for(k=0;k<2;k++)
                temp.a[i][j]+=(a.a[i][k]*b.a[k][j]);
            temp.a[i][j]%=MOD;
        }
        return temp;
}


matrix power(LL n)
{
     matrix temp,s;
        temp.a[0][1]=temp.a[1][0]=0;
        temp.a[0][0]=temp.a[1][1]=1;
        while(n!=0)
        {
          if(n%2!=0)
            temp=multi(temp,s);
        s=multi(s,s);
        n=n/2;
        }
        return temp;
}

int main()
{
    LL n;
    while(~scanf("%lld",&n)&&(n!=-1))
    {

        matrix t=power(n);
        cout<<t.a[1][0]<<endl;
    }
    return 0;
}