Question

Write a program mexp that multiplies a square matrix by itself a specified number of times、mexp takes a single argument, which is the path to a file containing a square (k × k) matrix M and a non-negative exponent n. It computes M and prints the result Note that the size of the matrix is not known statically. You ust use malloc to allocate space for the matrix once you obtain its size from the input file. Tocompute M". it is sufficient to multiply M by itself n _ 1 times. That is,AP-M×M×M Naturally, a different strategy is needed for r Input format The first line of the input file contains an integer k. This indicates the size of the matrix M, which has k rows and k columns. The next k lines in the input file contain k integers. These indicate the content of M.Each line corresponds to a row, beginning with the first (top) row The final line contains an integer n. This indicates the number of times M will be nultiplied by itself, n is guaranteed to be non-negative, but it may be 0 For example, an input file file.txt containing 1 2 3 4 5 6 7 89 indicates that sexp must compute Output format The output of mexp is the computed matrix M. Each row of M is printed on a separate line, beginning with the first (top) row. The items within a row are separated by spaces Using file.txt from above, /nexp filel.txt 30 36 42 66 81 96 102 126 150 This is done in C on command line in Linux

Answer 1

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#include <stdio.h>

#include <stdlib.h>

int main(int argc, char *argv[])

{

if (argc != 2)

{

printf("Usage: ./mexp file1.txt\n");

return 0;

}

FILE *fp;

fp = fopen(argv[1], "r");

if (fp == NULL)

{

printf("Unable to open file\n");

return 0;

}

int i, j, k, m, n, times;

fscanf(fp, "%d", &n);

int **M = (int **)malloc(n * sizeof(int *));

int **M1 = (int **)malloc(n * sizeof(int *));

int **res = (int **)malloc(n*sizeof(int *));

for (i = 0; i < n; i++){

M[i] = (int *)malloc(n * sizeof(int));

M1[i] = (int *)malloc(n * sizeof(int));

res[i] = (int *)malloc(n * sizeof(int));

}

for (i = 0; i < n; i++)

{

for (j = 0; j < n; j++)

{

fscanf(fp, "%d", &M[i][j]);

M1[i][j] = M[i][j];

}

}

fscanf(fp, "%d", &times);

for (m = 0; m < times-1; m++)

{

for (i = 0; i < n; i++)

{

for (j = 0; j < n; j++)

{

res[i][j] = 0;

for (k = 0; k < n; k++)

res[i][j] += M[i][k] * M1[k][j];

}

}

for(i=0;i<n; i++){

for(j=0;j<n; j++){

M1[i][j] = res[i][j];

}

}

}

for(i=0;i<n; i++){

for(j=0;j<n; j++){

printf("%d ", res[i][j]);

}

printf("\n");

}

fclose(fp);

return 0;

}