POJ 3233 Matrix Power Series 矩阵快速幂求A+A2+A3+…+Ak

题意 ：给出n k m 和一个n*n的矩阵A 求A + A² + A³ + … + A^k

参考http://blog.csdn.net/wangjian8006/article/details/7868864

构造矩阵很重要啊！！！ 弱菜不会啊

#include <cstdio>
#include <cstring>
const int mod = 10000;
const int maxn = 66;
struct Mat
{
    int a[maxn][maxn];
};
Mat A, B;
int n, m;
Mat get(Mat x, Mat y)
{
    Mat z;
    memset(z.a, 0, sizeof(z.a));
    for(int i = 1; i <= 2*n; i++)
        for(int j = 1; j <= 2*n; j++)
            for(int k = 1; k <= 2*n; k++)
            {
                z.a[i][j] += x.a[i][k]*y.a[k][j];
                z.a[i][j] %= m;
            }
    return z;
}
void Mat_pow(int x)
{
	//puts("s");
	if(x <= 0)
		return;
    while(x)
    {
        if(x&1)
            B = get(B, A);
        A = get(A, A);
        x >>= 1;
    }
}
int main()
{
	int x;
	while(scanf("%d %d %d", &n, &x, &m) != EOF)
	{
		memset(A.a, 0, sizeof(A.a));
		memset(B.a, 0, sizeof(B.a));
		for(int i = 1; i <= n; i++)
			for(int j = 1; j <= n; j++)
				scanf("%d", &A.a[i][j]);
		for(int i = 1; i <= n; i++)
			{
				A.a[i][i+n] = A.a[i+n][i+n] = 1;
				B.a[i][i] = B.a[i+n][i+n] = 1;
			}
		Mat_pow(x+1);
		for(int i = 1; i <= n; i++)
		{
			for(int j = 1; j <= n; j++)
			{
				if(i == j)
				{
					B.a[i][j+n] += m-1;
					B.a[i][j+n] %= m;
				}
				if(j-1)
					printf(" ");
				printf("%d", B.a[i][j+n]);
			}
			puts("");
		}
	}
	return 0;
}