* hello * * world ****** Whereas the phrase "the sky is falling"might be ****** .... the * sky * is * falling *********** Or *********** the sky is * falling ....
* hello * * world ****** Whereas the phrase "the sky is falling"might be ****** .... the * sky * is * falling *********** Or *********** the sky is * falling ....
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
Related questions
Question
Can you help me to write this program in C++.
Expert Solution
Step 1
C++ Code:
#include <iostream>
#include <vector>
using namespace std;
// using std::vector;
// using std::string;
// using std::cout;
// using std::cin;
// using std::endl;
int main() {
vector<string> strings;
cout << "Enter string: ";
for(;;) {
string s;
cin >> s;
strings.push_back(s);
if(getchar() == '\n')
break;
}
unsigned int n, i, j;
cout << "Enter size: ";
cin >> n;
// assuming strings.size() < n
unsigned int empty_lines_around_text((n - strings.size()) / 2);
cout << "empty_lines_around_text = " << empty_lines_around_text << endl;
// first horizontal row of stars
for(j = 0; j < n; ++j)
cout << '*';
cout << endl;
for(i = 1; i < empty_lines_around_text; ++i) {
cout << '*';
for(j = 1; j < n - 1; ++j) {
cout << ' ';
}
cout << '*' << endl;
}
//here we do the actual printing of the strings
for(i = 0; i < strings.size(); ++i) {
string s = strings[i];
// once again, assuming the size of each string is < n
unsigned int empty_chars_around_string((n - s.size()) / 2);
cout << '*';
for(j = 0; j < empty_chars_around_string; ++j)
cout << ' ';
cout << s;
for(j = empty_chars_around_string + s.size() + 1; j < n - 1; ++j)
cout << ' ';
cout << '*' << endl;
}
for(i = empty_lines_around_text + strings.size() + 1; i < n; ++i) {
cout << '*';
for(j = 1; j < n - 1; ++j) {
cout << ' ';
}
cout << '*' << endl;
}
// last horizontal line of '*' (we close the square)
for(j = 0; j < n; ++j)
cout << '*';
cout << endl;
return 0;
}
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