Counting Pixels

最新推荐文章于 2024-08-05 07:42:49 发布

G_rass

最新推荐文章于 2024-08-05 07:42:49 发布

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本文链接：https://blog.youkuaiyun.com/G_rass/article/details/45793933

本文介绍了一种计算在高分辨率屏幕上绘制圆形时点亮像素数量的方法，通过确定圆心坐标及半径，利用数学公式精确计算点亮的像素数。

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Description

Did you know that if you draw a circle that fills the screen on your 1080p high definition display, almost a million pixels are lit? That's a lot of pixels! But do you know exactly how many pixels are lit? Let's find out!

Assume that our display is set on a Cartesian grid where every pixel is a perfect unit square. For example, one pixel occupies the area of a square with corners (0,0) and (1,1). A circle can be drawn by specifying its center in grid coordinates and its radius. On our display, a pixel is lit if any part of is covered by the circle being drawn; pixels whose edge or corner are just touched by the circle, however, are not lit.

Your job is to compute the exact number of pixels that are lit when a circle with a given position and radius is drawn.

Input

The input consists of several test cases, each on a separate line. Each test case consists of three integers, x,y, and r(1≤x,y,r≤1,000,000), specifying respectively the center (x,y) and radius of the circle drawn. Input is followed by a single line with x = y = r = 0, which should not be processed.

Output

For each test case, output on a single line the number of pixels that are lit when the specified circle is drawn.

Assume that the entire circle will fit within the area of the display.

Sample Input

1 1 1
5 2 5
0 0 0

Sample Output

4
88

HINT

#include<stdio.h>
#include<math.h>
main()
{
long long a,b,r;
while(scanf("%lld%lld%lld",&a,&b,&r)!=EOF){
  if(a==0&&b==0&&r==0) continue;
  long long sum=0;
  double y;
  for(long long i=0;i<r;i++){
   y=sqrt(r*r-i*i);
   sum+=ceil(y);
  }
  printf("%lld\n",4*sum);
}
}

题解：

这里要注意r的范围很大，所以程序中多处用long long 类型（切不可遗漏任一处），另注意ceil的用法及其与floor的比对

#include<stdio.h>
#include<math.h>
 
intmain(void)
{
double number=123.54;
double down,up;
down=floor(number);
up=ceil(number);
printf("originalnumber%5.2lf\n",number);
printf("numberroundeddown%5.2lf\n",down);
printf("numberroundedup%5.2lf\n",up);
 
return0;
}

运行结果：

originalnumber123.54
numberroundeddown123.00
numberroundedup124.00