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

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