poj 3070 Fibonacci

矩阵快速幂

while(N)
 {
      if(N&1)
          res=res*A;
      n>>=1;
      A=A*A;
 }

#include<iostream>
using namespace std;
#deinfe mod 10000
struct matrix
{
    int m[2][2];
}ans,base;
matrix multi(matrix a, matrix b)
{
    matrix t;

    for(int i = 0; i < 2; i++)
    {
        for(int j = 0; j < 2; j++)
        {
            t.m[i][j] = 0;
            for(int pos = 0; pos < 2; pos++)
            {
                t.m[i][j] = (t.m[i][j] + a.m[i][pos] * a[pos][j]) % mod;
            }
        }
    }
    return t;
}
int fast(int n)
{

}

int main()
{
    freopen("input.txt","r",stdin);
    int n;
    while(scanf("%d",&n) && n != -1)
    {
        printf("%d
",fast());
    }
    return 0;
}