第10次落地时,共经过多少米?第10次反d多高?
1.程序分析:见下面注释
2.程序源代码:
#include "stdio.h"
#include "stdio.h"
main()
{
float sn=100.0,hn=sn/腔运2
int n
for(n=2n<=10n++)
{
sn=sn+2*hn/*第n次落地时共经过的米数*/
hn=hn/2/*第n次反跳高度*/
}
printf("the total of road is %f\n",sn)
printf("the tenth is %f meter\n",hn)
getch()
}
这个是WIN-TC的教程,如果你用C++编译器,不用打最后的getch()
#include <iostream>#include <math.h>
#define pi 3.1415926
#include <iostream>
#include <math.h>
#define pi 3.1415926
void main()
{
double radius = 0
double density = 0
std::cout <<"半径:"
std::cin >>radius
std::cout <<"密度慎闭睁:"
std::cin >>density
double area = 4 * pi * pow(radius, 2)
std::cout <<态迟 "表面积:" <<area <<std::endl
double volume = 4 * pi * pow(radius, 3)
std::cout <<"体积:" <<volume <<std::endl
double quality = density * volume
std::cout <<"宽岁质量:" <<quality <<std::endl
}
要制造那种效果只需要大约 30 行 Java 代卖戚橡码:
import javax.swing.*
import java.awt.*
import java.awt.geom.*
class RollingBall extends JPanel {
Ellipse2D.Float ball = new Ellipse2D.Float( -100, 100, 50, 50 )
public void paintComponent( Graphics g ) {
super.paintComponent( g )
Graphics2D g2 = ( Graphics2D ) g
// Draw the ball
g2.fill( ball )
// Draw the rotating ellipse by skewing the Device Space
double angdeg = // One rotation per ball's travelling over its perimeter
ball.x++ % ( Math.PI * ball.width ) / ( Math.PI * ball.width ) * 360
g2.rotate( Math.toRadians( angdeg ), ball.getCenterX( ), ball.getCenterY( ) )
g2.scale( .5, 1 )
g2.translate( ball.getCenterX( ), 0 )
g2.setColor( Color.gray )
g2.fill( ball )
}
public void roll( ) throws Exception {
while( true ) {
repaint( )
Thread.sleep( 8 )
中旁 }
}
public static void main( String[ ] args ) throws Exception {
JFrame f = new JFrame( )
RollingBall rb = new RollingBall( )
仔枝 f.setSize( 999, 185 )
f.getContentPane( ).add( rb )
f.setVisible( true )
rb.roll( )
}
}
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