Counting Pixels--计算几何

题目链接：http://acm.tzc.edu.cn/acmhome/problemdetail.do?&method=showdetail&id=3475

Counting Pixels

Time Limit(Common/Java):3000MS/9000MS     Memory Limit:65536KByte
Total Submit: 33            Accepted: 9

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(less than 30) test cases, each on a separate line. Each test case consists of three integers, x,
y, and r (1<= x, y, r <=1000,000), 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

分析：

这题一眼就可以看出方格的个数肯定是4的倍数，因此我们只需要找在1/4圆内有几个方格就可以了。这时应该采取极限的思想，只需要判断圆弧边上的方格是否在圆内就可以了。

代码如下：

#include <stdio.h>
#include <math.h>
int main()
{
    __int64 x, y, r;
    while(scanf("%I64d %I64d %I64d", &x, &y, &r) , x||y||r)
    {
        __int64 i, sum = 0;;
        for(i = 0; i < r; i++)
        {
            double d = sqrt(r*r-i*i);
            __int64 l = (__int64)d;
            if(l == d) sum += l;    //如果d是整数的话只需要加上l个方格就可以了
            else sum += l+1;    //如果d是小数的话需要加上l+1个方格（最边上的方格有一部分在圆内）
        }
        printf("%I64d\n", 4*sum);
    }
    return 0;
}