题目：在1个2n*2n的网格中间画1个直径为2n⑴的圆，问圆内部的格子和和圆相交的格子个数。
分析：计算几何。分别计算出圆和每一个整数横坐标的交点（xi，yi）。
ceil（yi）- ceil（yi⑴）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+1⑴e⑴2) - (int)(r+1⑴e⑴2)+1)<<2;
in += ((int)(r+1e⑴2))<<2;
l = r;
}
on += ((int)(l+1⑴e⑴0))<<2;
if (count ++) printf("
");
printf("In the case n = %d, %d cells contain segments of the circle.
",n,on);
printf("There are %d cells completely contained in the circle.
",in);
}
return 0;
}