CSUOJ 1011 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

Hint

---

思路：一个圆的大小是确定的，那么就将圆心放到原点，此时只需求第一象限的方块个数，然后乘4就行了

一开始最容易想到的是直接算方块左下角到原点的距离算出来，然后判断是否比半径小就ok了

#include<stdio.h>
#include<iostream>
using namespace std;
typedef long long ll;
int main()
{
	ll x, r, y;
	ll dis,num;
	while (~scanf("%lld%lld%lld", &x, &y, &r))
	{
		if (!x&&!y&&!r)
			break;
		ll R = r;
		num = 0;
		for (r; r - 1 >= 0; r--)
		{
			for (ll j = 0; j < R; j++)
			{
				dis = j*j + (r - 1)*(r - 1);
				if (dis <= R*R)
					num++;
				if (dis == R*R&&r - 1 >= 0)
					num--;
			}
		}
		printf("%lld\n", 4 * num);
	}
	return 0;
}
/**********************************************************************
	Problem: 1011
	User: leo6033
	Language: C++
	Result: TLE
**********************************************************************/

很快就敲好了，结果一交，TLE

后来仔细，观察了下，发现直接算对应横坐标的高度然后除一想上取整就ok了。

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<math.h>
using namespace std;
typedef long long ll;
int main()
{
	ll x, r, y;
	ll num;
	while (~scanf("%lld%lld%lld", &x, &y, &r))
	{
		if (!x&&!y&&!r)
			break;
		num = 0;
		for (ll i = 0; i < r; i++)
		{
			num += (ll)ceil(sqrt((double)((r*r) - (i*i))));
		}
		printf("%lld\n", 4 * num);
	}
	return 0;
}
/**********************************************************************
	Problem: 1011
	User: leo6033
	Language: C++
	Result: AC
	Time:120 ms
	Memory:2036 kb
**********************************************************************/