Write it's main. char space[3][100] = { "#.*.#******#****.****#.*.", "#***#******#*.**.*.*.#***", "#*.*#******#*********#*.*" }; int space_ind; for(space_ind = 0; space[0][space_ind];space_ind++) { if(space[0][space_ind] == '#') printf("#"); elseif (space[2][space_ind] == '.' ) continue; elseif(space[0][space_ind] == '.') printf("A"); elseif(space[1][space_ind] == '.') printf("I"); elseif(space[0][space_ind+1] == '.') printf("U"); elseif(space[1][space_ind+1] == '.') printf("O"); elseprintf("E"); if(space[0][space_ind] == '#' || space[2][space_ind] == '.' ) continue; else space_ind+=2; } return0;.

The provided code is a C program that prints a sequence of characters based on the contents of the…

Answered: { "#.*.#******#****.****#.*.",…

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Question

Write it's main.

char space[3][100] =
	{
	"#..#***#.*#..",
	"#*#***#.*...#**",
	"#.#****#******#.*"
	};

	int space_ind;

	for(space_ind = 0; space[0][space_ind];space_ind++)
	{
	if(space[0][space_ind] == '#')
	printf("#");
	elseif (space[2][space_ind] == '.' )
	continue;
	elseif(space[0][space_ind] == '.')
	printf("A");
	elseif(space[1][space_ind] == '.')
	printf("I");
	elseif(space[0][space_ind+1] == '.')
	printf("U");
	elseif(space[1][space_ind+1] == '.')
	printf("O");
	elseprintf("E");
	if(space[0][space_ind] == '#' \|\| space[2][space_ind] == '.' )
	continue;
	else
	space_ind+=2;

	}

	return0;.

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