Description

Given a n × n matrix A and a positive integer k, find the sum S=A+A2+A3++Ak.

Input

The input contains exactly one test case. The first line of input contains three positive integers n (n ≤ 30), k (k ≤ 109) and m (m

Output

Output the elements of S modulo m in the same way as A is given.

Sample Input

2 2 4
0 1
1 1

Sample Output

1 2
2 3

Solution

Bn+1=A0IIn+1=An0A+A2+A3+...+An+II

Bn+1B[0][1]I即可。

Code

include #include
#include
#include
using namespace std;
#define maxn 100
typedef long long ll;
struct Mat
{
int mat[maxn][maxn];//矩阵
int row, col;//矩阵行列数
};
Mat mod_mul(Mat a, Mat b, int p)//矩阵乘法
{
Mat ans;
ans.row = a.row;
ans.col = b.col;
memset(ans.mat, 0, sizeof(ans.mat));
for (int i = 1;i for (int j = 1;j for (int k = 1;k return ans;
}
Mat mod_pow(Mat a, int k, int p)//矩阵快速幂
{
Mat ans;
ans.row = a.row;
ans.col = a.col;
for (int i = 1;i for (int j = 1;j while (k)
{
if (k & 1)ans = mod_mul(ans, a, p);
a = mod_mul(a, a, p);
k >>= 1;
}
return ans;
}
int main()
{
int n, m, k;
while (~scanf("%d%d%d", &n, &k, &m))
{
Mat A;
A.row = A.col = 2 * n;
memset(A.mat, 0, sizeof(A.mat));
for (int i = 1; i for (int j = 1; j scanf("%d", &A.mat[i][j]);
for (int i = 1; i 1;//初始化单位矩阵
Mat B = mod_pow(A, k + 1, m);
for (int i = 1; i for (int j = n+1; j 2*n; j++)
{
if (i + n == j) printf("%d", ((B.mat[i][j] - 1) % m+m)%m);//((B.mat[i][j] - 1) % m+m)%m 避免结果为负数
else printf("%d", B.mat[i][j]);
printf("%c", j == 2 * n ? '\n' : ' ');
}

}
return 0;
}