POJ 3233 Matrix Power Series

Matrix Power Series

Time Limit: 3000MS		Memory Limit: 131072K
Total Submissions: 20210		Accepted: 8478

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

 

解法有多种。

1.利用分块矩阵

  思路比较简单，建立下面分块矩阵，

  各个分块矩阵都是m*m的，I表示单位矩阵，O表示零矩阵，可以将它变化为幂次形式，即，

 

#include <iostream>
#include <algorithm>
#include <map>
#include <vector>
#include <functional>
#include <string>
#include <cstring>
#include <queue>
#include <set>
#include <cmath>
#include <cstdio>
using namespace std;
#define IOS ios_base::sync_with_stdio(false)]
typedef long long LL;
const int INF=0x3f3f3f3f;

const int maxn=61;
int modnum, n,k;
typedef struct matrix{
    int v[maxn][maxn];
    void init(){memset(v,0,sizeof(v));}
}M;
M a,b,a2,d,r;
M mul_mod(const M &a,const M &b,int L,int m,int n)
{
    M c; c.init();
    for(int i=0;i<L;i++){
        for(int j=0;j<n;j++){
            for(int k=0;k<m;k++)//注意j，k范围
                c.v[i][j]=(c.v[i][j]+a.v[i][k]*b.v[k][j]%modnum)%modnum;
        }
    }
    return c;
}
M power(M x,int L,int p)
{
    M tmp; tmp.init();
    for(int i=0;i<L;i++)
        tmp.v[i][i]=1;
    while(p){
        if(p&1) tmp=mul_mod(x,tmp,L,L,L);
        x=mul_mod(x,x,L,L,L);
        p>>=1;
    }
    return tmp;
}
void init()
{
    b.init();
    for(int i=0;i<n;i++)
        b.v[i][i]=b.v[i][i+n]=1;
    for(int i=n;i<2*n;i++)
        for(int j=n;j<2*n;j++)
            b.v[i][j]=d.v[i%n][j%n]=a.v[i%n][j%n];
    a2=power(a,n,2);
    for(int i=n;i<2*n;i++)
        for(int j=0;j<n;j++)
            d.v[i][j]=a2.v[i%n][j];
}

int main()
{
    while(scanf("%d%d%d",&n,&k,&modnum)==3){
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                scanf("%d",&a.v[i][j]);
                a.v[i][j]%=modnum;
            }
        }
        if(k==1){
            for(int i=0;i<n;i++){
                for(int j=0;j<n;j++)
                    printf("%d ",a.v[i][j]);
                printf("\n");
            }
            continue;
        }
        init();
        r=power(b,2*n,k-1);
        r=mul_mod(r,d,2*n,2*n,n);
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++)
                printf("%d ",r.v[i][j]);
            printf("\n");
        }
    }
}