2021-06-05-优快云博客

该博客主要讨论了一个数学问题，即在1x1的网格上，将长度和宽度划分为50个部分，形成2500个单位网格。然后，给定n个半径为R的等径圆，求未被任何圆覆盖的单位网格数量。博客中提供了输入输出示例及解题代码，通过计算每个网格到圆心的距离来确定是否被覆盖，最终输出未被覆盖的网格总数。

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									Grids
									On an 1 × 1 grid,partition the length and width into 50 parts,forming 2500 unit grids with the same size.Now,given n circles with equal radius R on the grid.Find the total number of unit grids which are not covered by any of a circle.

A unit grid is covered by a circle,means that the distance between the center of the unit grid and the center of circle is less or equal R.

输入

There will be multiple test cases.Each line of input contains a Integer n (1≤n≤100) and a decimal R (0.01≤R≤0.1).Follows n lines,each line describe the center of circle i,include two decimal xi, yi (0≤xi, yi≤1).

输出

For each test case output a line,output the number of the points which are not covered.

样例输入

1 0.02
0.03 0.03
1 0.10
0.952423 0.042419

样例输出

2495
2454

提示

The picture for the first sample input.

题解：

#include<bits/stdc++.h>
#include<algorithm>
#include<queue>
using namespace std;
typedef unsigned long long ll;
const int N=5e5+10;
const ll mod= 1e9+7;
struct node
{
	double x,y;
}h[105];
double sum=0;
int main()
{
   string s;
   int n;
   double r;
   while(cin>>n>>r)
   {
   	r*=100;
   	for(int i=0;i<n;i++)
   	{
   		cin>>h[i].x>>h[i].y;
   		h[i].x*=100;
   		h[i].y*=100;
	   }
	   int sum=0;
	   for(double  i=1;i<=100;i+=2)
	   {
	   for(double j=1;j<=100;j+=2)
	   {
	   	int f=0;
	   	for(int v=0;v<n;v++)
	   	{
	   		if(sqrt((i*1.00000-h[v].x)*(i*1.0-h[v].x)+(j*1.0-h[v].y)*(j*1.0-h[v].y))<=r)
	   		{
	   			f=1;
	   			break;
			   }
		   }
		   if(f)
		   sum++;
	   }
}
cout<<2500-sum<<endl;
	} 
	return 0;
}