如何解决根据用户\ input绘制ASCII艺术风格的心脏 画一个 ASCII 艺术的心，填满空
Java程序应基于用户给出的输入N打印ASCII心。
信息：
- 要打印的字符：♡♥
- 心脏的 top 和 bottom 部分的打印可以分别进行
- 解决方案必须基于for循环
- N确定心脏的顶部：1. 顶部的倾斜外侧有N个心脏。 2. 顶部的扁平部分有N个心脏。 。3. 两个平坦部分之间的间隙为N个心。
示例：
我当前的代码：
Scanner scan = new Scanner(system.in);
System.out.println("Enter the number of hearts you want to print");
int userInput = scan.nextInt();
printTop(userInput);
}
public static void printTop(int userInput) {
String row = "";
int width = 2 * (userInput - 1) + 3 * userInput;
for (int height = 0; height < userInput; height++) {
for (int i = 0; i < userInput - 1; i++) {
row += "♡";
}
for (int i = 0; i < userInput; i++) {
row += "♥";
}
for (int i = 0; i < userInput; i++) {
row += "♡";
}
for (int i = 0; i < userInput; i++) {
row += "♥";
}
for (int i = 0; i < userInput - 1; i++) {
row += "♡";
}
row += "\n";
}
System.out.println(row);
}
想法：
问题：
- 我怎样才能使不同类型的心脏在每一行上完成其工作？
解决方法
画一个 ASCII 艺术的心，填满空
想象一个坐标平面并绘制一个由两个半圆和一个半菱形组成的图形。
图 1 和图 2：
图 1：
int n = 6;
for (int y = -n; y <= 2 * n; y++) {
for (int x = -2 * n; x <= 2 * n; x++)
if ((y <= 0 &&
((int) Math.sqrt((x+n)*(x+n) + y*y) <= n
|| (int) Math.sqrt((x-n)*(x-n) + y*y) <= n))
|| (y > 0 && Math.abs(x) <= 2 * n - y))
System.out.print("♥ ");
else
System.out.print("♡ ");
System.out.println();
}
输出 1：
♡ ♡ ♡ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♡ ♡ ♡ ♡ ♡ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♡ ♡ ♡
♡ ♡ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♡ ♡ ♡ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♡ ♡
♡ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♡ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♡
♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥
♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥
♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥
♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥
♡ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♡
♡ ♡ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♡ ♡
♡ ♡ ♡ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♡ ♡ ♡
♡ ♡ ♡ ♡ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♡ ♡ ♡ ♡
♡ ♡ ♡ ♡ ♡ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♡ ♡ ♡ ♡ ♡
♡ ♡ ♡ ♡ ♡ ♡ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♡ ♡ ♡ ♡ ♡ ♡
♡ ♡ ♡ ♡ ♡ ♡ ♡ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♡ ♡ ♡ ♡ ♡ ♡ ♡
♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡
♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡
♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♥ ♥ ♥ ♥ ♥ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡
♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♥ ♥ ♥ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡
♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♥ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡
图 2：
int n = 6;
for (int y = -n; y <= 2 * n; y++) {
for (int x = -2 * n; x <= 2 * n; x++)
if ((y <= 0 &&
((int) Math.sqrt((x+n)*(x+n) + y*y) == n
|| (int) Math.sqrt((x-n)*(x-n) + y*y) == n))
|| (y > 0 && Math.abs(x) == 2 * n - y))
System.out.print("♥ ");
else
System.out.print("♡ ");
System.out.println();
}
输出 2：
♡ ♡ ♡ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♡ ♡ ♡ ♡ ♡ ♥ ♥ ♥ ♥ ♥ ♥ ♥ ♡ ♡ ♡
♡ ♡ ♥ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♥ ♡ ♡ ♡ ♥ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♥ ♡ ♡
♡ ♥ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♥ ♡ ♥ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♥ ♡
♥ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♥ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♥
♥ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♥ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♥
♥ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♥ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♥
♥ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♥ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♥
♡ ♥ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♥ ♡
♡ ♡ ♥ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♥ ♡ ♡
♡ ♡ ♡ ♥ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♥ ♡ ♡ ♡
♡ ♡ ♡ ♡ ♥ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♥ ♡ ♡ ♡ ♡
♡ ♡ ♡ ♡ ♡ ♥ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♥ ♡ ♡ ♡ ♡ ♡
♡ ♡ ♡ ♡ ♡ ♡ ♥ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♥ ♡ ♡ ♡ ♡ ♡ ♡
♡ ♡ ♡ ♡ ♡ ♡ ♡ ♥ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♥ ♡ ♡ ♡ ♡ ♡ ♡ ♡
♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♥ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♥ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡
♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♥ ♡ ♡ ♡ ♡ ♡ ♥ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡
♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♥ ♡ ♡ ♡ ♥ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡
♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♥ ♡ ♥ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡
♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♥ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡
另见：
• HERE
• Print out an ASCII circle and axes with characters
很好的挑战
import java.util.Scanner;
public class A {
public static final char COLORED = '♥';
public static final char WHITE = '♡';
public static void printTop(int userInput) {
StringBuffer row = new StringBuffer();
int width = 2 * (userInput - 1) + 3 * userInput;
int outerHearts = userInput - 1; // white hearts at the outside
int heartsSection1 = userInput; // left colored section or complete colored line from the widthest point
int heartsSection2 = userInput; // right color section or 0 from the widthest point
int innerHearts = userInput; // white hearts between 2 colored sections
boolean upperPart = true;
while (heartsSection1 > 0) {
for (int i = 0; i < outerHearts; i++) {
row.append(WHITE);
}
for (int i = 0; i < heartsSection1; i++) {
row.append(COLORED);
}
for (int i = 0; i < innerHearts; i++) {
row.append(WHITE);
}
for (int i = 0; i < heartsSection2; i++) {
row.append(COLORED);
}
for (int i = 0; i < outerHearts; i++) {
row.append(WHITE);
}
row.append("\n");
if (upperPart) {
if (outerHearts > 0) {
outerHearts--;
innerHearts = Math.max(0,innerHearts - 2);
if (outerHearts == 0) {
heartsSection1 = width;
heartsSection2 = 0;
} else {
heartsSection1 += 2;
heartsSection2 += 2;
}
} else {
// line with only colored hearts. From now on the hearts reduce
upperPart = false;
}
}
if (!upperPart) {
outerHearts++;
heartsSection1 -= 2;
}
if (heartsSection1 <= 0)
break;
}
System.out.println(row.toString());
}
public static void main(String args[]) {
System.out.println("Enter the number of hearts you want to print");
Scanner scan = new Scanner(System.in);
int userInput = scan.nextInt();
printTop(userInput);
}
}
在我的输出中，这不是一个完美的矩形，因为白色的心比彩色的心要小。
,我用以下代码解决了任务：
import java.util.Scanner;
public class ChristmasHearts {
private int userInput;
private int height;
private int width;
String topHeart = "";
String middleHeart = "";
String bottomHeart = "";
public ChristmasHearts(int input) {
this.userInput = input;
}
public void draw() {
calcWidthAndHeight(this.userInput);
drawTopOfHeart();
drawBottomOfHeart();
}
private void drawTopOfHeart() {
int whites = 0;
int blacks = 1;
int center = 1;
int heightOfTopHeart = 1;
int isEvenCenterLessSubtractor = 2;
int isNotEvenCenterLessSubtractor = 1;
if (this.userInput == 1) {
for (int i = 0; i < heightOfTopHeart; i++) {
this.topHeart += "♥";
this.topHeart += "♡";
this.topHeart += "♥";
}
System.out.println(topHeart);
} else {
whites = this.userInput - 1;
blacks = this.userInput;
center = this.userInput;
heightOfTopHeart = this.userInput - 1;
for (int i = 0; i < heightOfTopHeart; i++) {
for (int j = 0; j < whites; j++) {
this.topHeart += "♡";
}
for (int j = 0; j < blacks; j++) {
this.topHeart += "♥";
}
for (int j = 0; j < center; j++) {
this.topHeart += "♡";
}
if (this.userInput >= 5 && center <= 0) {
if (isEven(this.userInput) && center == 0) {
for (int j = 0; j < blacks; j++) {
this.topHeart += "♥";
}
} else if (isEven(this.userInput) && center < 0) {
int rightSideBlacks = blacks - isEvenCenterLessSubtractor;
for (int j = 0; j < rightSideBlacks; j++) {
this.topHeart += "♥";
}
isEvenCenterLessSubtractor += 2;
} else if (!isEven(this.userInput) && center < 0) {
int rightSideBlacks = blacks - isNotEvenCenterLessSubtractor;
for (int j = 0; j < rightSideBlacks; j++) {
this.topHeart += "♥";
}
isNotEvenCenterLessSubtractor += 2;
} else {
int factorToSubtract = 2;
int rightSideBlacks = blacks - factorToSubtract;
for (int j = 0; j < rightSideBlacks; j++) {
this.topHeart += "♥";
}
factorToSubtract += 2;
}
} else {
for (int j = 0; j < blacks; j++) {
this.topHeart += "♥";
}
}
for (int j = 0; j < whites; j++) {
this.topHeart += "♡";
}
whites -= 1;
blacks += 2;
center -= 2;
if (this.userInput == 0) {
this.topHeart = "";
}
System.out.println(topHeart);
topHeart = "";
}
}
}
private void drawMiddelOfHeart() {
for (int j = 0; j < this.width; j++) {
this.middleHeart += "♥";
}
if (this.userInput == 0) {
this.middleHeart = "";
}
System.out.println(this.middleHeart);
middleHeart = "";
}
private void drawBottomOfHeart() {
int whites = 1;
int blacks = 1;
int heightOfBottomHeart = 1;
if (this.userInput == 1) {
for (int i = 0; i < heightOfBottomHeart; i++) {
this.bottomHeart += "♡";
this.bottomHeart += "♥";
this.bottomHeart += "♡";
}
System.out.println(bottomHeart);
} else {
heightOfBottomHeart = this.height - this.userInput;
whites = 1;
blacks = this.width - 2;
drawMiddelOfHeart();
for (int i = 0; i < heightOfBottomHeart; i++) {
for (int j = 0; j < whites; j++) {
this.bottomHeart += "♡";
}
for (int j = 0; j < blacks; j++) {
this.bottomHeart += "♥";
}
for (int j = 0; j < whites; j++) {
this.bottomHeart += "♡";
}
blacks -= 2;
whites += 1;
if (this.userInput == 0) {
this.bottomHeart = "";
}
System.out.println(bottomHeart);
bottomHeart = "";
}
}
}
private boolean isEven(int n) {
return n % 2 == 0;
}
private void calcWidthAndHeight(int n) {
this.width = 3;
this.height = 2;
for (int counter = 1; counter <= (n - 1); counter++) {
this.width += 5;
if (isEven(counter)) {
this.height += 4;
} else {
this.height += 3;
}
}
}
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int input = scanner.nextInt();
scanner.close();
ChristmasHearts heart = new ChristmasHearts(input);
heart.draw();
}
}
,您可以使用 java.awt 包将这些字符 ♥♡ 绘制为黑白图片：
然后你可以打印一个二维像素数组作为组成这张图片的字符：
♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡
♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡
♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡
♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡
♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡
♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡
♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡
♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡
♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡
♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡
♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡
♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡
♡ ♡ ♡ ♡ ♡ ♡ ♡ ♡
♡ ♡ ♡ ♡ ♡ ♡
♡ ♡ ♡ ♡
♡ ♡
public static void main(String[] args) {
int[][] hearts = stringToPixels("♥♡",new Font(Font.SERIF,Font.PLAIN,22),0);
// output
for (int[] row : hearts) {
StringBuilder sb = new StringBuilder();
for (int i : row)
sb.append(i != 0 ? "♡ " : " ");
if (sb.indexOf("♡") > -1)
System.out.println(sb);
}
}
/**
* @param str source string to draw
* @param f font to render text
* @param pd padding as a precaution,in most cases 0
* @return the 2D array of pixels that make up this string
*/
static int[][] stringToPixels(String str,Font f,int pd) {
FontRenderContext ctx = new FontRenderContext(f.getTransform(),false,false);
Rectangle bounds = f.getStringBounds(str,ctx).getBounds();
int width = bounds.width + pd * 2;
int height = bounds.height + pd * 2;
BufferedImage image =
new BufferedImage(width,height,BufferedImage.TYPE_BYTE_BINARY);
Graphics2D graphics = (Graphics2D) image.getGraphics();
graphics.setFont(f);
graphics.drawString(str,pd + bounds.x,pd - bounds.y);
//ImageIO.write(image,"png",new File("str.png"));
int[][] pcx = new int[height][width];
for (int i = 0; i < height; i++)
for (int j = 0; j < width; j++)
if (image.getRGB(j,i) == 0xFFFFFFFF)
pcx[i][j] = 1;
return pcx;
}
最简单的方法是为这条线建立一个char[]，用空心填充它，然后用白色心填充两侧，然后计算并用白色心填充中心。
结果是非常简洁和快速的代码，但中心计算可能看起来很复杂，虽然它是一个简单的公式。
public static void printHeart(int n) {
printHeart(n,'\u2665','\u2661');
}
public static void printHeart(int n,char black,char white) {
if (n <= 0)
return;
int width = 5 * n - 2;
char[] buf = new char[width];
for (int i = 1 - n; i < (width + 1) / 2; i++) {
int outer = Math.abs(i),center = 2 - n - 2 * i;
Arrays.fill(buf,black);
Arrays.fill(buf,outer,white);
Arrays.fill(buf,width - outer,width,white);
if (center > 0)
Arrays.fill(buf,(width - center) / 2,(width + center) / 2,white);
System.out.println(buf);
}
}
请注意代码如何不需要单独的部分来打印上半部分和下半部分。这是通过从负数到正数迭代 i 实现的技巧，例如对于 n=4，i 从 -3 迭代到 +8，两者都包含。
测试
由于我浏览器的字体没有使用等宽的心形，结果不好，所以下面的测试使用的是正则字符。
for (int n = 0; n <= 6; n++)
printHeart(n,'X','.');
输出
X.X
.X.
.XX..XX.
XXXXXXXX
.XXXXXX.
..XXXX..
...XX...
..XXX...XXX..
.XXXXX.XXXXX.
XXXXXXXXXXXXX
.XXXXXXXXXXX.
..XXXXXXXXX..
...XXXXXXX...
....XXXXX....
.....XXX.....
......X......
...XXXX....XXXX...
..XXXXXX..XXXXXX..
.XXXXXXXXXXXXXXXX.
XXXXXXXXXXXXXXXXXX
.XXXXXXXXXXXXXXXX.
..XXXXXXXXXXXXXX..
...XXXXXXXXXXXX...
....XXXXXXXXXX....
.....XXXXXXXX.....
......XXXXXX......
.......XXXX.......
........XX........
....XXXXX.....XXXXX....
...XXXXXXX...XXXXXXX...
..XXXXXXXXX.XXXXXXXXX..
.XXXXXXXXXXXXXXXXXXXXX.
XXXXXXXXXXXXXXXXXXXXXXX
.XXXXXXXXXXXXXXXXXXXXX.
..XXXXXXXXXXXXXXXXXXX..
...XXXXXXXXXXXXXXXXX...
....XXXXXXXXXXXXXXX....
.....XXXXXXXXXXXXX.....
......XXXXXXXXXXX......
.......XXXXXXXXX.......
........XXXXXXX........
.........XXXXX.........
..........XXX..........
...........X...........
.....XXXXXX......XXXXXX.....
....XXXXXXXX....XXXXXXXX....
...XXXXXXXXXX..XXXXXXXXXX...
..XXXXXXXXXXXXXXXXXXXXXXXX..
.XXXXXXXXXXXXXXXXXXXXXXXXXX.
XXXXXXXXXXXXXXXXXXXXXXXXXXXX
.XXXXXXXXXXXXXXXXXXXXXXXXXX.
..XXXXXXXXXXXXXXXXXXXXXXXX..
...XXXXXXXXXXXXXXXXXXXXXX...
....XXXXXXXXXXXXXXXXXXXX....
.....XXXXXXXXXXXXXXXXXX.....
......XXXXXXXXXXXXXXXX......
.......XXXXXXXXXXXXXX.......
........XXXXXXXXXXXX........
.........XXXXXXXXXX.........
..........XXXXXXXX..........
...........XXXXXX...........
............XXXX............
.............XX.............
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