nyoj_299_Matrix Power Series_矩阵快速幂

Matrix Power Series

时间限制：1000 ms  |  内存限制：65535 KB

难度：4

 

描述

Given a n × n matrix A and a positive integer k, find the sum S = A + A² + A³ + … + A^k.

 

输入

The input contains exactly one test case. The first line of input contains three positive integers n (n ≤ 30), k (k ≤ 10^9) and m (m < 10^4). Then follow n lines each containing n nonnegative integers below 32,768, giving A’s elements in row-major order.

输出

Output the elements of S modulo m in the same way as A is given.

样例输入

2 2 4
0 1
1 1

样例输出

1 2
2 3

来源

POJ Monthly

上传者

张云聪

#include <iostream>
#include <cstdio>

using namespace std;

int M=1000007;

struct Matrix{
    long int line,column;
    long int m[40][40];
};
struct Matr{
    long int line,column;
    long int m[70][70];
    Matr(Matrix x){
        line =x.line*2;
        column=x.column*2;
        for(int i=0;i<x.line*2;i++){
            for(int j=0;j<x.column;j++){
                m[i][j]=x.m[i%x.line][j];
            }
        }
        for(int i=0;i<x.line;i++){
            for(int j=x.column;j<x.column*2;j++){
                m[i][j]=0;
            }
        }
        for(int i=x.line;i<x.line*2;i++){
            for(int j=x.column;j<x.column*2;j++){
                if(i==j){
                    m[i][j]=1;
                }else{
                    m[i][j]=0;
                }
            }
        }
    }
};

Matr mult(Matr a,Matr b){
    Matr ans(a);
    ans.line=a.line;
    ans.column=b.column;
    //ans=inist(ans,0);
    for(int i=0;i<ans.line;i++){
        for(int j=0;j<ans.column;j++){
            ans.m[i][j]=0;
            for(int k=0;k<ans.column;k++){
                ans.m[i][j]+=(a.m[i][k]*b.m[k][j]);
                ans.m[i][j]%=M;
            }
        }
    }
    return ans;
}

Matr fast_matrix(Matr x,int n){
    Matr an(x),tmp(x);
    for(int i=0;i<x.line;i++){
        for(int j=0;j<x.column;j++){
            an.m[i][j]=x.m[i+x.line/2][j];
        }
    }
    an.line/=2;
    while(n){
        if(n%2!=0){
            an=mult(an,tmp);
        }
        tmp=mult(tmp,tmp);
        n>>=1;
    }
    return an;
}

int main()
{
    int n,m,k;
    Matrix a;
    scanf("%d %d %d",&n,&k,&m);
    M=m;
    a.line=n;
    a.column=n;
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++){
            scanf("%d",&a.m[i][j]);
        }
    }
    Matr ans(a);
    Matr ans2=fast_matrix(ans,k-1);
    for(int i=0;i<ans2.line;i++){
        for(int j=0;j<ans2.column/2;j++){
            printf("%d ",ans2.m[i][j]);
        }
        printf("
");
    }
    return 0;
}