POJ 3233-Matrix Power Series( S = A + A^2 + A^3 + … + A^k 矩阵快速幂取模)

Matrix Power Series

Time Limit: 3000MS		Memory Limit: 131072K
Total Submissions: 20309		Accepted: 8524

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 < 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

题目意思：

已知n，k，m，S = A + A2 + A3 + … + Ak. 求：S%m

解题思路：

模板题，快速幂取模+二分优化。

补充一个关于等比序列分治的结论：

对于
Sn=(A^1+A^2+A^3+……+A^(n-1)+A^n) mod p
当n为偶数的时候
s
=(1+A^(n/2))* (A^1+A^2+A^3+……+A^(n/2)) =(1+A^(n/2))*S[n/2]
当n为奇数的时候
s
=(1+A^((n-1)/2+1))* (A^1+A^2+A^3+……+A^(n-1)/2)+A^((n-1)/2+1]
=(1+A^((n-1)/2+1))* S[(n-1)/2] + A^((n-1)/2+1) =(1+A^(n/2+1))*S[n/2] + A^(n/2+1)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <algorithm>
#define INF 0xfffffff
using namespace std;
const long long MAXN=81;
long long n,mod;
struct Mat
{
long long m[MAXN][MAXN];
};
Mat a,per;
void init()
{
long long i,j;
for(i=0; i<n; ++i)
for(j=0; j<n; ++j)
{
cin>>a.m[i][j];
a.m[i][j]%=mod;
per.m[i][j]=(i==j);
}
}

Mat mul(Mat A,Mat B)
{
Mat ans;long long i,j,k;
for(i=0; i<n; i++)
for(j=0; j<n; j++)
{
ans.m[i][j]=0;
for(k=0; k<n; k++)
ans.m[i][j]+=(A.m[i][k]*B.m[k][j]);
ans.m[i][j]%=mod;
}
return ans;
}
Mat power(long long k)
{
Mat p,ans=per;
p=a;
while(k)
{
if(k&1)
{
ans=mul(ans,p);
--k;
}
else
{
k/=2;
p=mul(p,p);
}
}
return ans;
}

Mat add(Mat a,Mat b)
{
Mat c;long long i,j;
for(i=0; i<n; ++i)
for(j=0; j<n; ++j)
c.m[i][j]=(a.m[i][j]+b.m[i][j])%mod;
return c;
}
Mat sum(long long k)
{
if(k==1) return a;
Mat temp,b;
temp=sum(k/2);
if(k&1)
{
b=power(k/2+1);
temp=add(temp,mul(temp,b));
temp=add(temp,b);
}
else
{
b=power(k/2);
temp=add(temp,mul(temp,b));
}
return temp;
}

int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
long long i,j,k;
while(cin>>n>>k>>mod)
{
init();
Mat ans=sum(k);
for(i=0; i<n; ++i)
{
for(j=0; j<n-1; ++j)
cout<<ans.m[i][j]<<" ";
cout<<ans.m[i][j]<<endl;
}
}
return 0;
}