Matrix Power Series

poj 3233

題目大意

給你一個矩陣A，讓你求$S=A+A^2+A^3+\dots +A^k$

輸入樣例

2 2 4 0 1 1 1

輸出樣例

1 2 2 3

$n?30,k?10^9,m<10^4,a\in A,a?32768$

解題思路

如果直接計算k遍會超時
看到矩陣我們考率矩陣乘法（doge）
我們設矩陣$\left[\begin{array}{cc}{A}^{i?1}& S_{i?2}\end{array}\right]$（A,S均為$n?n$的矩陣，$S_i=A+A^1+A^2\dots +A^i$）
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為單位矩陣
這樣就可以快速冪了

代碼

#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//矩陣乘法{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)//快速冪 {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;//建初始矩陣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; }

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