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

1 1 1
5 2 5
0 0 0

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