poj 3233 Matrix Power Series
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H - Matrix Power Series
Given a n × n matrix A and a positive integer k, find the sumS = A + A2 + A3 + … + Ak.
Input
The input contains exactly one test case. The first line of input contains three positive integersn (n ≤ 30), k (k ≤ 109) and m (m < 104). Then follown lines each containing n nonnegative integers below 32,768, givingA’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
题意:已知一个n*n的矩阵A,和一个正整数k,求S = A + A2 + A3 + … + Ak。
思路:矩阵快速幂。首先我们知道 A^x 可以用矩阵快速幂求出来(具体可见poj 3070)。其次可以对k进行二分,每次将规模减半,分k为奇偶两种情况,如当k = 6和k = 7时有:
k = 10 有: S(10) = ( A^1+A^2+A^3+A^4+ A^5 ) + A^5 * ( A^1+A^2+A^3+A^4+A^5 ) = S(5) + A^5 * S(5)
k = 5 有: S(5) = ( A^1+A^2 ) + A^3 + A^3 * ( A^1+A^2 ) = S(2) + A^3 + A^3 * S(2)
k = 2 有 : S(2) = A^1 + A^2 = S(1) + A^1 * S(1)
二分 递归 矩阵 快速幂
#include <cstdio>
#include <cstring>
using namespace std ;
struct node
{
int a[50][50] ;
};
int n,k,m;
node mul(node p,node q)//矩阵乘法
{
node s;
int i,j,k;
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
s.a[i][j]=0 ;
for(k=0;k<n;k++)
{
s.a[i][j]=(s.a[i][j]+p.a[i][k]*q.a[k][j])%m;
}
}
}
return s ;
}
node add(node p,node q)//矩阵加法
{
int i,j;
node s;
for(i=0;i<n;i++)
for(j=0;j<n;j++)
s.a[i][j]=(p.a[i][j]+q.a[i][j])%m ;
return s ;
}
node pow(node p,int k)//矩阵快速幂
{
if(k==1)
return p;
node s=pow(p,k/2);
s=mul(s,s);
if(k%2)
s=mul(s,p);
return s;
}
node f(node p,int k)//二分递归
{
if(k==1)
return p;
node s=f(p,k/2),q,temp;
int i,j;
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
if(i==j)
temp.a[i][j]=1;
else
temp.a[i][j]=0;
}
}////标准矩阵*任何矩阵都为1
if(k%2)//奇数
{
q=pow(p,k/2+1) ;
s=add(q,mul(s,add(q,temp))) ;
}
else
{
q=pow(p,k/2);
s=mul(s,add(q,temp));
}
return s ;
}
int main()
{
int i,j;
node p,s;
while(scanf("%d%d%d",&n,&k,&m)!=EOF)
{
for(i=0;i<n;i++)
for(j=0;j<n;j++)
scanf("%d",&p.a[i][j]);
s=f(p,k);
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
if(j==n-1)
printf("%d\n",s.a[i][j]);
else
printf("%d ",s.a[i][j]);
}
}
}
return 0;
}
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