CSU1011: Counting Pixels

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 it 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

Sample Output

4

88

题意：给定圆心和半径，要你找这个圆覆盖了多少的矩形。由于圆是中心对称，所以考虑四分之一的圆。
那么怎么想？考虑右上方的四分之圆，如果一个一个矩形被覆盖，那么这个矩形的左下角的点到圆心的距离一定小于半径，可以自己画下图理解，如果这个矩形在圆内，那么这个矩形以下的一列都会被圆覆盖，所以我们考虑离圆心最远
的每个矩形，不断的向右向下走，直到它运动到圆心的水平线下。最后乘以4就是答案，不懂可以看代码和画图理解一下，应该不难。

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <math.h>

using namespace std;

const int maxn=1005;

typedef long long LL;

int x,y,r;

int main()

{

    while(scanf("%d %d %d",&x,&y,&r)!=EOF)

    {

        if(x==0&&y==0&&r==0)

            return 0;

        else

        {

            LL ans=0;

            int i=r-1,j=0;

            LL temp=r*r;

            while(j<r)

            {

                if(i*i+j*j<temp)

                    ans+=(i+1);

                else

                {

                    i--;

                    continue;

                }

                j++;

            }

            printf("%lld\n",ans*4);

        }

    }

    return 0;

}

/**********************************************************************

	Problem: 1011

	User: therang

	Language: C++

	Result: AC

	Time:28 ms

	Memory:2024 kb

**********************************************************************/