csuoj-1011-Counting Pixels

Posted 2020-09-09 znt

题目：

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

分析：
1，根据对称性，只求左上角区域即可；
2，从1到r-1，求出对应直角三角形的高，如果刚好为整数，就加上h,否则加上h + 1，最后再加上r，再乘以4。
代码：

#include<iostream>
#include<cmath>
using namespace std;
int main(){
    cin.sync_with_stdio(false);
    long long x,y,r;
    while(cin >> x >> y >> r){
        if(x == 0 && y == 0 && r == 0)
            break;
        else{
            long long ans = r;
            for(long long i = 1;i <= r - 1;i++){
                long long h = (long long)sqrt((long double)r * r - i * i);
                if(h * h + i * i == r * r) ans += h;
                else ans += h + 1;
            }
            cout << 4 * ans << endl;
        }
    }
    return 0;
}