用 * 打印 X
Printin a X with *
我想用 * 打印 X,我已经完成了 X 的左侧,但我不知道如何打印另一侧 (flip/mirror)。
如果你 运行 这段代码,它将只打印 (X) 的左侧,现在我想打印 (X) 的右侧?那么我应该怎么做才能使用星号(*)完成(X)呢?谢谢你们。
我想知道是否可以这样做?(我是编程新手)
#include <iostream>
// Expected output pattern:
//
// * *
// * *
// * *
// *
// * *
// * *
// * *
using namespace std;
int main() {
cout << "Printing X with star(*)" << endl;
cout << endl;
int i;
int p;
for (i = 1; i <= 10; i++) {
for (int j = 1; j <= 10; j++) {
if (j > i) break;
cout << " ";
cout << "\t";
}
cout << "\t\t\t\t";
for (p = 1; p <= 10; p++) {
cout << "*";
}
cout << endl;
}
for (i = 10; i >= 1; i--) {
for (int j = 1; j <= 10; j++) {
if (j > i) break;
cout << " ";
cout << "\t";
}
cout << "\t\t\t\t";
for (p = 1; p <= 10; p++) {
cout << "*";
}
cout << endl;
}
return 0;
}
你走在正确的轨道上,要在右手边做你必须在每行上打印更多 **** 除了你已经完成的。将 X 的每一行打印为打印一些 **** 然后一些 spaces 然后更多 **** 并在每次接近时减少 spaces 的数量可能会有所帮助交叉点。那有意义吗?这可能会帮助你走得更远 (. = space):
*......*
.*....*
..*..*
...**
等等
这是您可以到达那里的众多方式之一:
int main()
{
int size = 8;
int spacesBefore;
int spacesBetween = size;
int numStars = 1;
// Top half:
int i, j;
for ( i = 0; i < size/2; i++ ) {
spacesBetween = size - ( 2 * ( i + 1 ) );
spacesBefore = i;
for ( j = 0; j < spacesBefore; j++ ) // before
cout << " ";
for ( j = 0; j < numStars; j++ ) // * left
cout << "*";
for ( j = 0; j < spacesBetween; j++ ) // between
cout << " ";
for ( j = 0; j < numStars; j++ ) // * right
cout << "*";
cout << endl;
}
// bottom half, do the same kind of thing but changing the spacings
// ...
}
好的,谢谢所有帮助过我的人,我在将近 6 个小时后找到了我想要的答案,答案如下:
#include <iostream>
using namespace std;
int main() {
cout << "Printing X with stars" << endl;
cout << endl;
int i;
int p;
int k;
int s;
int count = 72;
for (i = 1; i <= 10; i++) {
for (int j = 1; j <= 10; j++) {
if (j > i) break;
cout << " ";
cout << "\t";
}
cout << "\t\t\t\t";
for (p = 1; p <= 10; p++) {
cout << "* ";
}
for (k=1; k<=count; k++){
cout << " ";
}
count-=8;
for (s=1; s<=10; s++){
cout << "* ";
}
cout << endl;
}
count = 0;
for (i = 10; i >= 1; i--) {
for (int j = 1; j <= 10; j++) {
if (j > i) break;
cout << " ";
cout << "\t";
}
cout << "\t\t\t\t";
for (p = 1; p <= 10; p++) {
cout << "* ";
}
for (k=1; k<=count; k++) {
cout << " ";
}
count +=8;
for (s=1; s<=10; s++){
cout << "* ";
}
cout << endl;
if (count == 80) break;
}
return 0;
}
我想用 * 打印 X,我已经完成了 X 的左侧,但我不知道如何打印另一侧 (flip/mirror)。 如果你 运行 这段代码,它将只打印 (X) 的左侧,现在我想打印 (X) 的右侧?那么我应该怎么做才能使用星号(*)完成(X)呢?谢谢你们。 我想知道是否可以这样做?(我是编程新手)
#include <iostream>
// Expected output pattern:
//
// * *
// * *
// * *
// *
// * *
// * *
// * *
using namespace std;
int main() {
cout << "Printing X with star(*)" << endl;
cout << endl;
int i;
int p;
for (i = 1; i <= 10; i++) {
for (int j = 1; j <= 10; j++) {
if (j > i) break;
cout << " ";
cout << "\t";
}
cout << "\t\t\t\t";
for (p = 1; p <= 10; p++) {
cout << "*";
}
cout << endl;
}
for (i = 10; i >= 1; i--) {
for (int j = 1; j <= 10; j++) {
if (j > i) break;
cout << " ";
cout << "\t";
}
cout << "\t\t\t\t";
for (p = 1; p <= 10; p++) {
cout << "*";
}
cout << endl;
}
return 0;
}
你走在正确的轨道上,要在右手边做你必须在每行上打印更多 **** 除了你已经完成的。将 X 的每一行打印为打印一些 **** 然后一些 spaces 然后更多 **** 并在每次接近时减少 spaces 的数量可能会有所帮助交叉点。那有意义吗?这可能会帮助你走得更远 (. = space):
*......*
.*....*
..*..*
...**
等等
这是您可以到达那里的众多方式之一:
int main()
{
int size = 8;
int spacesBefore;
int spacesBetween = size;
int numStars = 1;
// Top half:
int i, j;
for ( i = 0; i < size/2; i++ ) {
spacesBetween = size - ( 2 * ( i + 1 ) );
spacesBefore = i;
for ( j = 0; j < spacesBefore; j++ ) // before
cout << " ";
for ( j = 0; j < numStars; j++ ) // * left
cout << "*";
for ( j = 0; j < spacesBetween; j++ ) // between
cout << " ";
for ( j = 0; j < numStars; j++ ) // * right
cout << "*";
cout << endl;
}
// bottom half, do the same kind of thing but changing the spacings
// ...
}
好的,谢谢所有帮助过我的人,我在将近 6 个小时后找到了我想要的答案,答案如下:
#include <iostream>
using namespace std;
int main() {
cout << "Printing X with stars" << endl;
cout << endl;
int i;
int p;
int k;
int s;
int count = 72;
for (i = 1; i <= 10; i++) {
for (int j = 1; j <= 10; j++) {
if (j > i) break;
cout << " ";
cout << "\t";
}
cout << "\t\t\t\t";
for (p = 1; p <= 10; p++) {
cout << "* ";
}
for (k=1; k<=count; k++){
cout << " ";
}
count-=8;
for (s=1; s<=10; s++){
cout << "* ";
}
cout << endl;
}
count = 0;
for (i = 10; i >= 1; i--) {
for (int j = 1; j <= 10; j++) {
if (j > i) break;
cout << " ";
cout << "\t";
}
cout << "\t\t\t\t";
for (p = 1; p <= 10; p++) {
cout << "* ";
}
for (k=1; k<=count; k++) {
cout << " ";
}
count +=8;
for (s=1; s<=10; s++){
cout << "* ";
}
cout << endl;
if (count == 80) break;
}
return 0;
}