POJ3233 Matrix Power Series

题意

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Matrix Power Series

Time Limit: 3000MS		Memory Limit: 131072K
Total Submissions: 29541		Accepted: 11975

Description

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

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 < 10⁴). Then follow n lines each containing n nonnegative integers below 32,768, giving A’s elements in row-major order.

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

Source

POJ Monthly--2007.06.03, Huang, Jinsong

分析

跟整数的分治差不多。

\[\sum_{i=1}^kA^i=\left\{\begin{array}{} (I+A^{k/2})\sum_{i=1}^{k/2}A^i&,&k\equiv 0\ (\bmod 2) \\ A+(A+A^{(k+1)/2})\sum_{i=1}^{(k-1)/2}A^i&,&k\equiv 1\ (\bmod 2) \end{array}\right. \]

时间复杂度\(O(n^3 \log k)\)

#include<iostream>
#include<cstring>
#define rg register
#define il inline
#define co const
template<class T>il T read(){
    rg T data=0,w=1;rg char ch=getchar();
    while(!isdigit(ch)) {if(ch=='-') w=-1;ch=getchar();}
    while(isdigit(ch)) data=data*10+ch-'0',ch=getchar();
    return data*w;
}
template<class T>il T read(rg T&x) {return x=read<T>();}
typedef long long ll;

co int N=30;
struct M{int a[N][N];};
int n,k,m;
M add(co M&x,co M&y){
	M ans;
	for(int i=0;i<n;++i)
		for(int j=0;j<n;++j)	
		ans.a[i][j]=(x.a[i][j]+y.a[i][j])%m;
	return ans;
}
M mul(co M&x,co M&y){
	static M ans;
	memset(ans.a,0,sizeof ans.a);
	for(int i=0;i<n;++i)
		for(int j=0;j<n;++j)
			for(int k=0;k<n;++k)
				ans.a[i][j]=(x.a[i][k]*y.a[k][j]%m+ans.a[i][j])%m;
	return ans;
}
M ksm(M x,int k){
	static M ans;
	memset(ans.a,0,sizeof ans.a);
	for(int i=0;i<n;++i) ans.a[i][i]=1;
	for(;k;x=mul(x,x),k>>=1)
		if(k&1) ans=mul(ans,x);
	return ans;
}
M get(co M&x,int k){
	if(k==1) return x;
	M y=ksm(x,(k+1)>>1);
	M z=get(x,k>>1);
	return k&1?add(x,mul(add(x,y),z)):mul(add(ksm(x,0),y),z);
}
int main(){
//	freopen(".in","r",stdin),freopen(".out","w",stdout);
	M x;
	read(n),read(k),read(m);
	for(int i=0;i<n;++i)
		for(int j=0;j<n;++j) x.a[i][j]=read<int>()%m;
	x=get(x,k);
	for(int i=0;i<n;++i,puts(""))
		for(int j=0;j<n;++j) printf("%d ",x.a[i][j]);
	return 0;
}