UVa 356 - Square Pegs And Round Holes

题目：在一个2n*2n的网格中间画一个直径为2n-1的圆，问圆内部的格子以及和圆相交的格子个数。

分析：计算几何。分别计算出圆和每个整数横坐标的交点（xi，yi）。

            ceil（yi）- ceil（yi-1）即为每列的相交格子个数，floor（yi）即为每列圆内格子个数。

说明：注意精度(⊙_⊙)。

#include <algorithm>#include <iostream>#include <cstdlib>#include <cstring>#include <cstdio>#include <cmath>using namespace std;int main(){int n,in,on,count = 0;while (cin >> n) {in = 0,on = 0;double r,l = n-0.5;for (int i = 1 ; i < n ; ++ i) {r = sqrt((n-0.5)*(n-0.5)-i*i);on += ((int)(l+1-1e-12) - (int)(r+1-1e-12)+1)<<2;in += ((int)(r+1e-12))<<2;l = r;}on += ((int)(l+1-1e-10))<<2;if (count ++) printf("\n");printf("In the case n = %d, %d cells contain segments of the circle.\n",n,on);printf("There are %d cells completely contained in the circle.\n",in);}    return 0;}