poj 3233 Matrix Power Series(矩阵二分，快速幂）

Matrix Power Series

Time Limit: 3000MS		Memory Limit: 131072K
Total Submissions: 15739		Accepted: 6724

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

当k为奇数时

Sk = A + A2 + A3 + … + Ak

=(1+Ak/2)*(A + A2 + A3 + … + Ak/2 )+{Ak}

=(1+Ak/2)*(Sk/2 )+{Ak}

当k为偶数时

Sk = A + A2 + A3 + … + Ak

=(1+Ak/2)*(A + A2 + A3 + … + Ak/2 )+{Ak}

=(1+Ak/2)*(Sk/2 )

就可以二分递归求Sk

代码：

//829ms
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
int n,m;
struct matrix
{
int ma[50][50];
} a;
matrix multi(matrix x,matrix y)//矩阵相乘
{
matrix ans;
memset(ans.ma,0,sizeof(ans.ma));
for(int i=1; i<=n; i++)
{
for(int j=1; j<=n; j++)
{
if(x.ma[i][j])//稀疏矩阵优化
for(int k=1; k<=n; k++)
{
ans.ma[i][k]=(ans.ma[i][k]+x.ma[i][j]*y.ma[j][k])%m;
}
}
}
return ans;
}
matrix add(matrix x,matrix y)//矩阵相加
{
for(int i=1; i<=n; i++)
for(int j=1; j<=n; j++)
x.ma[i][j]=(x.ma[i][j]+y.ma[i][j])%m;
return x;
}
matrix pow(matrix a,int m)
{
matrix ans;
for(int i=1; i<=n; i++) //单位矩阵
{
for(int j=1; j<=n; j++)
{
if(i==j)
ans.ma[i][j]=1;
else
ans.ma[i][j]=0;
}
}
while(m)//矩阵快速幂
{
if(m&1)
{
ans=multi(ans,a);
}
a=multi(a,a);
m=(m>>1);
}
return ans;
}
matrix solve(matrix x,int k)//递归求Sk
{
if(k==1)
return x;
matrix ans;
for(int i=1; i<=n; i++) //单位矩阵
{
for(int j=1; j<=n; j++)
{
if(i==j)
ans.ma[i][j]=1;
else
ans.ma[i][j]=0;
}
}
ans=add(ans,pow(x,k/2));
ans=multi(ans,solve(x,k/2));
if(k&1)
ans=add(ans,pow(x,k));
return ans;
}
int main()
{
int k;
while(~scanf("%d%d%d",&n,&k,&m))
{
matrix ans,a;
for(int i=1; i<=n; i++)
{
for(int j=1; j<=n; j++)
scanf("%d",&a.ma[i][j]);
}
ans=solve(a,k);
for(int i=1; i<=n; i++)
{
for(int j=1; j<n; j++)
printf("%d ",ans.ma[i][j]);
printf("%d\n",ans.ma[i]
);
}
}
return 0;
}