Matrix Power Series

poj 3233

题目大意

给你一个矩阵A，让你求$S=A+A^2+A^3+\dots +A^k$

输入样例

2 2 4
0 1
1 1

输出样例

1 2
2 3

$n⩽30,k⩽10^9,m<10^4,a\in A,a⩽32768$

解题思路

如果直接计算k遍会超时
看到矩阵我们考率矩阵乘法（doge）
我们设矩阵$\left[\begin{array}{cc}{A}^{i-1}& S_{i-2}\end{array}\right]$（A,S均为$n\ast n$的矩阵，$S_i=A+A^1+A^2\dots +A^i$）
那么有$\left[\begin{array}{cc}{A}^i& s_{i-1}\end{array}\right]=\left[\begin{array}{cc}{A}^{i-1}& s_{i-2}\end{array}\right]\times \left[\begin{array}{cc}A& 1\\ 0& 1\end{array}\right]$
注：1为单位矩阵
这样就可以快速幂了

代码

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define wyc mod
ll n, k, mod;
struct matrix
{ll n, m, a[100][100];matrix operator *(const matrix &b) const//矩阵乘法{matrix c;c.n = n;c.m = b.m;for (ll i = 1; i <= c.n; ++i)for (ll j = 1; j <= c.m; ++j)c.a[i][j] = 0;for (ll i = 1; i <= n; ++i)for (ll k = 1; k <= m; ++k)for (ll j = 1; j <= c.m; ++j)c.a[i][j] = (c.a[i][j] + a[i][k] * b.a[k][j] % wyc) % wyc;return c;}
}A, B;
void Counting(ll g)//快速幂
{while(g){if (g & 1) A = A * B;B = B * B;g >>= 1; }
}
int main()
{scanf("%lld%lld%lld", &n, &k, &mod);A.n = n;A.m = n * 2;B.n = n * 2;B.m = n * 2; for (ll i = 1; i <= n; ++i)for (ll j = 1; j <= n; ++j){scanf("%lld", &A.a[i][j]);B.a[i][j] = A.a[i][j];}for (ll i = 1; i <= n; ++i)B.a[i][i + n] = B.a[i + n][i + n] = 1;//建初始矩阵Counting(k);for (ll i = 1; i <= n; ++i){for (ll j = 1; j <= n; ++j)printf("%lld ", A.a[i][j + n] % wyc);putchar(10);}return 0;
}