我正在尝试通过以下代码制作此形状。我很困惑如何在打印明星之前打印第二行，倒数第二行而不跳过它并打印额外的空间。一旦确定了下半部分，当星星向外扩展时，代码是否与上半部相似？我在c和r之间尝试了几种代码组合，但我一直坚持我目前的情况。
---------------------- //row 0
* *| //row 1
* * * *| //row 2
* * * * * *|
* * * * * * * *|
* * * * * * * * * *|
* * * * * * * * * * *|
* * * * * * * * * *|
* * * * * * * *|
* * * * * *|
* * * *|
* *|
----------------------
#include <iostream>
using std::cout; using std::cin; using std::endl;
int main() {
cout << "Enter a positive odd number less than 40: ";
int num = 0;
int z = 1;
for (int a = 0; a < 3; ++a)
{
cin >> num;
if (num < 38 && num > 0 && num % 2 == 1)
{
cout << "Thank you!" << endl << endl;
for (int r = 0; r < num; ++r) //outer loop/rows
{
for (int c = 0; c < num; ++c) //inner loop/columns
{
if (r == 0) cout << "--"; //top of square
else if (c >= r + r - c && c < num - 1)
cout << " ";
//else if (c == num - 1) cout << "*|";
else if (r == num - 1) cout << "--"; //bottom of square
else if (c == num - 1) cout << "*|"; //right side of square
else if (r > c) cout << "* ";
}
cout << endl;
}
break;
}
else cout << "Please enter a positve odd number that is less than 40!" << endl;
}
cout << endl;
}
答案 0 :(得分：3)
我刚刚接受了两个变量left=0＆amp; right=num-1并增加left＆amp;在right或r<=num/2我打印col <= left之后，我将col >=right或*推迟#include <iostream>
using std::cout; using std::cin; using std::endl;
int main() {
cout << "Enter a positive odd number less than 40: ";
int num = 0;
int z = 1;
for (int a = 0; a < 3; ++a)
{
cin >> num;
if (num < 38 && num > 0 && num % 2 == 1)
{
cout << "Thank you!" << endl << endl;
int left=0,right=num-1;
//for printing top line
for(int i = 0; i < num; i++) cout<<"- ";
cout<<"-"<<endl;
for (int r = 0; r < num; ++r) //outer loop/rows
{
//printing columns
for(int c = 0; c < num; c++)
{
if(c <= left || c >= right)
cout<<"* ";
else
cout<<" ";
}
if(r >= num/2) //checking for half of the rows
{
left--;right++;
}
else
{
left++;right--;
}
cout<<"|"<<endl;
}
//for printing last additional line
for(int i = 0; i < num; i++) cout<<"- ";
cout<<"-"<<endl;
break;
}
else cout << "Please enter a positve odd number that is less than 40!" << endl;
}
cout << endl;
}
，直到sudo locale-gen pt_BR
。
我希望这很容易理解
这是代码：
python -c "import locale; locale.setlocale(locale.LC_ALL, 'pt_BR')"
答案 1 :(得分：1)
这种方法是数学方式。
此外，它在边缘绘制一个带有加号的完整帧。
试一试。
#include <iostream>
#include <cmath>
using std::cout; using std::cin; using std::endl;
int main() {
cout << "Enter a positive odd number less than 40: ";
int num = 0;
int z = 1;
for (int a = 0; a < 3; ++a) {
cin >> num;
if (num < 40 && num > 0 && num % 2 == 1) {
cout << "Thank you!" << endl << endl;
int center = ceil(num / 2.0);
for (int r = 0; r <= num+1; ++r) { //outer loop/rows
for (int c = 0; c <= num+1; ++c) { //inner loop/columns
if (r == 0 || r == num+1) {
if (c == 0 || c == num+1)
cout << "+"; // corner
else
//top or botton of square between corners
if (c == center)
cout << "-";
else
cout << "--";
}
else if (c == 0 || c == num+1) {
cout << "|"; // left or right frame
} else {
// inner part
if ((center-std::abs(center-r)) >= center-std::abs(center-c))
if (c < center)
cout << "* ";
else if (c > center)
cout << " *";
else
cout << "*";
else
if (c == center)
cout << " ";
else
cout << " ";
}
}
cout << endl;
}
} else
cout << "Please enter a positve odd number that is less than 40!" << endl;
}
cout << endl;
}
答案 2 :(得分：1)
另一种方式（通过更多用户输入检查）：
#include <iostream>
#include <string>
#include <limits>
#include <sstream>
using std::cout;
using std::cin;
using std::string;
const auto ssmax = std::numeric_limits<std::streamsize>::max();
const int max_dim = 40;
const int max_iter = 3;
int main() {
cout << "Enter a positive odd number less than " << max_dim << ": ";
int num = 0, counter = 0;
while ( counter < max_iter ) {
cin >> num;
if ( cin.eof() )
break;
if ( cin.fail() ) {
cout << "Please, enter a number!\n";
cin.clear();
cin.ignore(ssmax,'\n');
}
if ( num < max_dim && num > 0 && num % 2 ) {
cout << "Thank you!\n\n";
//top line
string line(num * 2, '-');
cout << line << '\n';
for ( int r = 0, border = num - 1; r < num; ++r ) {
cout << '*';
for ( int c = 1; c < num; ++c ) {
if ( (c > r && c < border) || (c < r && c > border) )
cout << " ";
else
cout << " *";
}
// right border
cout << "|" << '\n';
--border;
}
//bottom line
cout << line << '\n';
++counter;
} else {
cout << "Please, enter a positive odd number that is less than 40!\n";
}
}
cout << std::endl;
}
或者我最喜欢的：
// top line
string line = string(num * 2, '-') + '\n';
cout << line;
// inside lines
int r = 0, border = ( num - 1 ) * 2;
string inside = string(border + 1, ' ') + "|\n";
// top
while ( r < border ) {
inside[r] = '*';
inside[border] = '*';
r += 2;
border -= 2;
cout << inside;
}
// center line
inside[r] = '*';
cout << inside;
// bottom
while ( border > 0 ) {
inside[r] = ' ';
inside[border] = ' ';
r += 2;
border -= 2;
cout << inside;
}
//bottom line
cout << line;