1064: 矩阵乘法

问题描述 给定一个N阶矩阵A，输出A的M次幂（M是非负整数）

例如：

A  =

1  2

3  4

A的2次幂

7  10

15  22

输入格式

第一行是一个正整数N、M（1< =N< =30, 0< =M< =5），表示矩阵A的阶数和要求的幂数

接下来N行，每行N个绝对值不超过10的非负整数，描述矩阵A的值

输出格式

输出共N行，每行N个整数，表示A的M次幂所对应的矩阵。相邻的数之间用一个空格隔开

样例输入

2  2

1  2

3  4

样例输出

7  10

15  22

import java.util.Scanner;

public class Main {


    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n,m;
        while(in.hasNext()) {
            //in
            n = in.nextInt();
            m = in.nextInt();
            int [][]A = new int[n][n];
            for (int i = 0; i < n; i++)
                for (int j = 0; j < n; j++)
                    A[i][j] = in.nextInt();
            //deal
            for (int i = 0; i < n && m==0; i++)
                for (int j = 0; j < n; j++) {
                    if (i==j)
                        A[i][j]=1;
                    else
                        A[i][j]=0;
                }

            int [][]B=A;
            int [][]C =A;
            while(m>1) {
                B = new int[n][n];
                m--;
                for (int i = 0; i < n; i++)
                    for (int j = 0; j < n; j++) {
                        for (int j2 = 0; j2 < n; j2++)
                            B[i][j] += C[i][j2]*A[j2][j];
                    }
                C=B;
            }

            //out
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++)
                    System.out.print(B[i][j]+" ");
                System.out.println();
            }
        }
    }

}
//C++
#include<iostream>
#include<stdio.h>
//#include<stdlib.h>
#include<string.h>
//#include<string>
//#include<math.h>
#include<algorithm>
#define inttochar(x) ('0'+x)
#define chartoint(x) (x-'0')
using namespace std;

int main()
{
    int n,m;
    while(cin>>n>>m){
      int A[n][n];
      for(int i=0;i<n;++i){
        for(int j=0;j<n;++j){
          cin>>A[i][j];
        }
      }
     //deal
            for (int i = 0; i < n && m==0; i++)
                for (int j = 0; j < n; j++) {
                    if (i==j)
                        A[i][j]=1;
                    else
                        A[i][j]=0;
                }

            int B[n][n];
            int C[n][n] ;
            for(int i=0;i<n;++i){
        for(int j=0;j<n;++j){
          B[i][j] = A[i][j];
          C[i][j] = A[i][j];
        }
      }
            while(m>1) {
        memset(B,0,sizeof(B));
                m--;
                for (int i = 0; i < n; i++)
                    for (int j = 0; j < n; j++)
                        for (int j2 = 0; j2 < n; j2++)
                            B[i][j] += C[i][j2]*A[j2][j];

                for(int i=0;i<n;++i){
        for(int j=0;j<n;++j){
          C[i][j] = B[i][j];
        }
      }

    }

            //out
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++)
                    cout<<B[i][j]<<" ";
                cout<<endl;
    }
    }
    return 0;
}