CF 392A Blocked Points - c++编程基础

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Blocked Points time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

Imagine you have an infinite 2D plane with Cartesian coordinate system. Some of the integral points are blocked, and others are not. Two integral points A and B on the plane are 4-connected if and only if:

the Euclidean distance between A and B is one unit and neither A nor B is blocked;
or there is some integral point C, such that A is 4-connected with C, and C is 4-connected with B.
Let's assume that the plane doesn't contain blocked points. Consider all the integral points of the plane whose Euclidean distance from the origin is no more than n, we'll name these points special. Chubby Yang wants to get the following property: no special point is 4-connected to some non-special point. To get the property she can pick some integral points of the plane and make them blocked. What is the minimum number of points she needs to pick
Input
The first line contains an integer n (0 ≤ n ≤ 4 107).

Output
Print a single integer ― the minimum number of points that should be blocked.
Sample test(s) input
```
1
```
output
```
4
```
input
```
2
```
output
```
8
```
input
```
3
```
output
```
16
```
暴力搜索，从第一象限的i=0，j=n开始搜索到i=n，j=0，找出每一个符合条件的点，最后*4即为所求。 PS:队友给力！
```
#include
    
     
long long n,m,sum;
bool dis(long long x,long long y)
{
    return x*x+y*y<=m;
}
int main()
{
    while(scanf("%lld",&n)!=EOF)
    {
        sum=0;
        m=n*n;
        bool flag;
        int pos=n;
        if(!n){printf("1\n");continue;}
        for(long long i=0;i
     
      =0;j--)//j=pos,每次从pos这个点开始往下搜，防止搜索不必要的点。次循环的条件是此点的左面或者上面必须有一个点在圆外 { if((i*i+j*j)<=m) { flag=true; sum++; pos=j; } } } printf("%d\n",sum*4); } return 0; } 
     
    
```