POJ 3070 矩阵快速幂

题意：求菲波那切数列的第n项。

分析：矩阵快速幂。

右边的矩阵为a0 ,a1，，，

然后求乘一次，就进一位，求第n项，就是矩阵的n次方后，再乘以b矩阵后的第一行的第一列。

#include <cstdio>
#include <cstring>

using namespace std;

typedef long long ll;

const int maxn = 101;
const int maxm = 101;

struct Matrix {
    int n,m;
    int a[maxn][maxm];

    void clear() {
        n = m = 0;
        memset(a,0,sizeof(a));
    }

    Matrix operator +  (const Matrix &b) const {
        Matrix tmp;
        tmp.n = n;
        tmp.m = m;

        for(int i=0;i<n;++i)
            for(int j=0;j<m;++j)
                tmp.a[i][j] = a[i][j] + b.a[i][j];

        return tmp;
    }

    Matrix operator - (const Matrix &b) const {
        Matrix tmp;
        tmp.n = n;
        tmp.m = m;

        for(int i=0;i<n;++i)
            for(int j=0;j<m;++j)
                tmp.a[i][j] = a[i][j] - b.a[i][j];

        return tmp;
    }

    Matrix operator * (const Matrix & b) const {
        Matrix tmp;
        tmp.clear();
        tmp.n = n;
        tmp.m = b.m;

        for(int i=0;i<n;++i)
            for(int j=0;j<b.m;++j)
                for(int k=0;k<m;++k)
                    tmp.a[i][j] +=(a[i][k]*b.a[k][j])%10000;

        return tmp;
    }

    Matrix operator ^ (const int& k) const {
        Matrix tmp,t = *this;

        int p = k;
        tmp.clear();

        tmp.m = tmp.n = this->n;

        for(int i=0;i<n;++i)
            tmp.a[i][i] = 1;

        while(p) {
            if(p&1) tmp = tmp*t;
            p>>=1;
            t = t*t;
        }
        return tmp;
      }
};

int main(int argc, char const *argv[])
{
    int n;
    while(true) {
        scanf("%d",&n);
        if(n==-1) break;
        Matrix f;
        f.clear();
        f.n = f.m = 2;
        f.a[0][0] = 0;
        f.a[0][1] = 1;
        f.a[1][0] = 1;
        f.a[1][1] = 1;

        f = f^n;
        Matrix x;
        x.clear();
        x.n = 2;
        x.m = 1;
        x.a[0][0] = 0;
        x.a[1][0] = 1;
        f = f*x;

        printf("%d\n", f.a[0][0]);

    }
    return 0;
}