用快速幂运算求斐波那契，时间复杂度降到O（logn）

#include <vector>
#include <iostream>
using namespace std;

typedef vector<int> vec;
typedef vector<vec> mat;
typedef long long ll;

const int M = 10000;
ll n;
mat mul(mat &A,mat &B)
{
    int m = A.size(), n = B[0].size(), k = B.size();
    mat C(m,vec(n));
    for(int i=0;i<m;i++)
    {
        for(int j=0;j<n;j++)
        {
            for(int counter=0;counter<k;counter++)
            {
                C[i][j] = (C[i][j]+A[i][counter]*B[counter][j])%M;
            }

        }
    }
    return C;
}

mat pow(mat A,ll n)
{
    mat B(A.size(),vec(A.size()));
    for(int i=0;i<A.size();i++)
    {
        B[i][i] = 1;
    }
    while(n>0)
    {
        if(n&1) B = mul(B,A);
        A = mul(A,A);
        n >>= 1;
    }
    return B;
}

void solve()
{
    mat A(2,vec(2));
    A[0][0] = 1; A[0][1] = 1;
    A[1][0] = 1; A[1][1] = 0;
    A = pow(A,n);
    cout<<"fib "<<n<<" is :"<<A[1][0];
}
int main()
{
    cin>>n;
    solve();
    return 0;
}