第二次作业
problem #1
第一题要求做一个由相同大小矩形砖块搭建而成的金字塔。砖块尺寸和底层砖块数由用户指定。要求是金字塔必须坐落在屏幕底部中央，而且每一层比下一层恰少一块砖。做出来应该是这个效果：
我的想法是利用两个for循环嵌套在一起实现。代码如下：
/*
* File: Pyramid.java
* ------------------
* This program is a stub for the Pyramid problem, which draws
* a brick pyramid.
*/
import acm.program.*;
import acm.graphics.*;
public class Pyramid extends GraphicsProgram {
public void run() {
int brickWidth;
int brickHeight;
int bricksInBase;
int windowWidth=getWidth();//获取窗口宽度
int windowHeight=getHeight();//获取窗口高度
println("This program will draw a pyramid of bricks in a window");
brickWidth=readInt("Enter brickWidth:");
brickHeight=readInt("Enter brickHeight:");
bricksInBase=readInt("Enter bricksInBase:");
for(int i=0;i<bricksInBase;i++){ //从底部开始绘制第i+1行
int r=bricksInBase-i; //r表示第i+1行的砖块数
int y=windowHeight-(i+1)*brickHeight; //确定第i+1行砖块的纵坐标位置
for(int j=0;j<r;j++){ //绘制第i+1行第j+1块砖块
int x=(windowWidth-(bricksInBase-i)*brickWidth)/2+j*brickWidth; //确定这一块的横坐标位置
GRect Brick= new GRect(x,y,brickWidth,brickHeight);
add(Brick);
}
}
}
}
problem #2
第二题要我们画出如下图所示的彩虹：
要求是，彩虹不能碰到窗口底部。彩虹的顶点必须在窗口中可见。
题目实现起来非常直白，因此有一点儿乏味：
/*
* File: Rainbow.java
* ------------------
* This program is a stub for the Rainbow problem, which displays
* a rainbow by adding consecutively smaller circles to the canvas.
*/
import java.awt.Color;
import acm.program.*;
import acm.graphics.*;
public class Rainbow extends GraphicsProgram {
public void run() {
int windowWidth=getWidth();
int windowHeight=getHeight();
GRect sky = new GRect(0,0,windowWidth,windowHeight);
sky.setColor(Color.CYAN);
sky.setFilled(true);
sky.setFillColor(Color.CYAN);
add(sky);
GOval circle1= new GOval(-143,27,1040,1040);
circle1.setFilled(true);
circle1.setFillColor(Color.RED);
add(circle1);
GOval circle2= new GOval(-123,47,1000,1000);
circle2.setFilled(true);
circle2.setFillColor(Color.ORANGE);
add(circle2);
GOval circle3= new GOval(-103,67,960,960);
circle3.setFilled(true);
circle3.setFillColor(Color.YELLOW);
add(circle3);
GOval circle4= new GOval(-83,87,920,920);
circle4.setFilled(true);
circle4.setFillColor(Color.GREEN);
add(circle4);
GOval circle5= new GOval(-63,107,880,880);
circle5.setFilled(true);
circle5.setFillColor(Color.BLUE);
add(circle5);
GOval circle6= new GOval(-43,127,840,840);
circle6.setFilled(true);
circle6.setFillColor(Color.MAGENTA);
add(circle6);
GOval circle7= new GOval(-23,147,800,800);
circle7.setFilled(true);
circle7.setFillColor(Color.CYAN);
add(circle7);
}
}
problem #3
如图，要求做一个这样的带文本的图。要求是整个图在窗口中居中。文本在各自的方框中居中。也是一道没啥好说的题，比较无趣。
/*
* File: GraphicsHierarchy.java
* ----------------------------
* This program is a stub for the GraphicsHierarchy problem, which
* draws a partial diagram of the acm.graphics hierarchy.
*/
import acm.program.*;
import acm.graphics.*;
public class GraphicsHierarchy extends GraphicsProgram {
public void run() {
int windowWidth=getWidth();
int windowHeight=getHeight();
int boxWidth=readInt("Enter class box ");
int boxHeight=readInt("Enter class box height:");
int x1,x2,x3,x4,x5,y1,y2;
x1=windowWidth/2;
y1=windowHeight/3;
y2=windowHeight*2/3;
x2=windowWidth/8;
x3=windowWidth*3/8;
x4=windowWidth*5/8;
x5=windowWidth*7/8;
GLine line1=new GLine(x1,y1,x2,y2);
add(line1);
GLine line2=new GLine(x1,y1,x3,y2);
add(line2);
GLine line3=new GLine(x1,y1,x4,y2);
add(line3);
GLine line4=new GLine(x1,y1,x5,y2);
add(line4);
GRect box1=new GRect(x1-boxWidth/2,y1-boxHeight,boxWidth,boxHeight);
add(box1);
GRect box2=new GRect(x2-boxWidth/2,y2,boxWidth,boxHeight);
add(box2);
GRect box3=new GRect(x3-boxWidth/2,y2,boxWidth,boxHeight);
add(box3);
GRect box4=new GRect(x4-boxWidth/2,y2,boxWidth,boxHeight);
add(box4);
GRect box5=new GRect(x5-boxWidth/2,y2,boxWidth,boxHeight);
add(box5);
GLabel label1=new GLabel("GObject",x1,y1);
int a1=(int)label1.getWidth();
int b1=(int)label1.getAscent();
label1.move(-a1/2,-boxHeight/2+b1/2);
add(label1);
GLabel label2=new GLabel("GLabel",x2,y2);
int a2=(int)label2.getWidth();
int b2=(int)label2.getAscent();
label2.move(-a2/2,boxHeight/2+b2/2);
add(label2);
GLabel label3=new GLabel("GLine",x3,y2);
int a3=(int)label3.getWidth();
int b3=(int)label3.getAscent();
label3.move(-a3/2,boxHeight/2+b3/2);
add(label3);
GLabel label4=new GLabel("GOval",x4,y2);
int a4=(int)label4.getWidth();
int b4=(int)label4.getAscent();
label4.move(-a4/2,boxHeight/2+b4/2);
add(label4);
GLabel label5=new GLabel("GRect",x5,y2);
int a5=(int)label5.getWidth();
int b5=(int)label5.getAscent();
label5.move(-a5/2,boxHeight/2+b5/2);
add(label5);
}
}
problem #4
此题要求用户输入一元二次方程的系数a b c 然后给出解。若无实根，则提示用户。
/*
* File: Quadratic.java
* --------------------
* This program is a stub for the Quadratic problem, which finds the
* roots of the quadratic equation.
*/
import acm.program.*;
public class Quadratic extends ConsoleProgram {
public void run() {
println("Enter coefficients for the quadratic equation:");
double a=readInt("a:");
double b=readInt("b:");
double c=readInt("c:");
double x=b*b-4*a*c;
if(x<0){
println("The equation has no real solutions!");
}
else{
double y=Math.sqrt(x);
double p=(-b+y)*0.5/a;
double q=(-b-y)*0.5/a;
println("The first solution is "+p);
println("The second solution is "+q);
}
}
}
problem #5
提示用户输入一串正整数，从中找出最大的和最小的，并使用用户指定的数作为输入完毕的标示。
/*
* File: FindRange.java
* --------------------
* This program is a stub for the FindRange problem, which finds the
* smallest and largest values in a list of integers.
*/
import acm.program.*;
public class FindRange extends ConsoleProgram {
public void run() {
int sentinel=readInt("Enter a integer as sentinel:");
println("This program finds the smallest and largest integers in a list. Enter values,one per line,using a "+sentinel+"to signal the end of the list");
int a=readInt("?");
int largest=0;
int smallest=1000000;
if(a==sentinel){
println("No values have been entered!");
}
else{
while (a!=sentinel){
if(a>largest){
largest=a;
}
if(a<smallest){
smallest=a;
}
a=readInt("?");
}
println("The smallest value is "+smallest);
println("The largest value is "+largest);
}
}
}
problem #6
最后一题是角谷猜想的一个简易验算的程序。提示用户输入一个正整数，要求显示出验算步骤和步骤数。
/*
* File: Hailstone.java
* --------------------
* This program is a stub for the Hailstone problem, which computes
* Hailstone sequence described in Assignment #2.
*/
import acm.program.*;
public class Hailstone extends ConsoleProgram {
public void run() {
println("This program computes Hailstone sequences.");
int a=readInt("Enter a number:");
int count=0;
while (a!=1){
if(a%2==0){
a=a/2;
println(2*a+" is even,so I take half = "+a);
count=count+1;
}
else{
a=3*a+1;
println((a-1)/3+" is odd,so I take 3n+1 ="+a);
count=count+1;
}
}
println("The process took "+count+" steps to reach 1");
}
}