Poj 3070 Fibonacci (Rapid matrix power)

Address: poj 3070

With this question, I learned to use the matrix's fast power to quickly calculate the Fibonacci number.

According to the previous formula, the number of the 1st columns in the 2nd rows and the 2nd columns in the 1st rows is the nth Fibonacci number. Therefore, you can construct a matrix and evaluate the power.

The Code is as follows:

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#include <set>#include <algorithm>using namespace std;const int mod=1e4;struct matrix{    int ma[3][3];}init, res;matrix Mult(matrix x, matrix y){    matrix tmp;    int i, j, k;    for(i=0;i<2;i++)    {        for(j=0;j<2;j++)        {            tmp.ma[i][j]=0;            for(k=0;k<2;k++)            {                tmp.ma[i][j]=(tmp.ma[i][j]+x.ma[i][k]*y.ma[k][j])%mod;            }        }    }    return tmp;}matrix Pow(matrix x, int k){    int i, j;    matrix tmp;    for(i=0;i<2;i++)    {        for(j=0;j<2;j++)        {            tmp.ma[i][j]=(i==j);        }    }    while(k)    {        if(k&1) tmp=Mult(tmp,x);        x=Mult(x,x);        k>>=1;    }    return tmp;}int main(){    int k;    while(scanf("%d",&k)!=EOF&&k>=0)    {        init.ma[0][0]=1;        init.ma[0][1]=1;        init.ma[1][0]=1;        init.ma[1][1]=0;        res=Pow(init,k);        printf("%d\n",res.ma[0][1]);    }    return 0;}

Poj 3070 Fibonacci (Rapid matrix power)