POJ 3233 Matrix Power Series（矩阵快速幂）

Matrix Power Series

Time Limit: 3000MS Memory Limit: 131072K

Total Submissions: 19338 Accepted: 8161

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

可以找到递推关系 : s[k]=s[k-1]+A^k;

然后构造矩阵，利用矩阵快速幂

具体见代码

#include <iostream>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>

using namespace std;
int n,k;
int m;
struct Node
{
    int a[65][65];

};
Node multiply(Node a,Node b)
{
    Node c;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            c.a[i][j]=0;
            for(int k=1;k<=n;k++)
            {
                (c.a[i][j]+=(a.a[i][k]*b.a[k][j])%m)%=m;
             }
        }
    }
    return c;
}
Node quick(Node a,int x)
{
    Node c;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
        c.a[i][j]=(i==j?1:0);
    for(x;x>0;x>>=1)
    {
        if(x&1)
            c=multiply(c,a);
        a=multiply(a,a);
     }
    return c;
}
int main()
{
   while( scanf("%d%d%d",&n,&k,&m)!=EOF)
   {
    Node a;Node b;Node c;
    memset(a.a,0,sizeof(a.a));
    memset(b.a,0,sizeof(b.a));
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
        {
            scanf("%d",&a.a[i][j+n]);
            b.a[i+n][j+n]=a.a[i][j+n];
        }
    for(int i=1;i<=n;i++)
    {
        b.a[i][i]=1;
        b.a[i+n][i]=1;
    }
    n=n*2;
    c=multiply(a,quick(b,k));
    for(int i=1;i<=n/2;i++)
        for(int j=1;j<=n/2;j++)
           if(j==n/2)printf("%d\n",c.a[i][j]);
           else printf("%d ",c.a[i][j]);
   }
    return 0;
}