【题解】poj3090 Visible Lattice Points（同洛谷P2158 [SDOI2008]仪仗队）欧拉函数

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#欧拉函数

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该博客介绍了一道编程题（poj3090/SDOI2008仪仗队），讨论了在第一象限中从原点可视的格点问题。题目要求计算在0到N范围内，与原点连线不经过其他格点的点的数量。解题关键在于理解可见点的条件（x, y > 0, gcd(x, y) = 1）并使用欧拉函数在O(N)时间内求解。给出的样例展示了不同N值下的可视点数。" 102958376,7814940,"C++ STL 容器详解：string, vector, deque, stack, queue, list, set, multiset, map, multimap

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题目链接

Description

A lattice point (x, y) in the first quadrant (x and y are integers greater than or equal to 0), other than the origin, is visible from the origin if the line from (0, 0) to (x, y) does not pass through any other lattice point. For example, the point (4, 2) is not visible since the line from the origin passes through (2, 1). The figure below shows the points (x, y) with 0 ≤ x, y ≤ 5 with lines from the origin to the visible points.

这里写图片描述
Write a program which, given a value for the size, N, computes the number of visible points (x, y) with 0 ≤ x, y ≤ N.

Input

The first line of input contains a single integer C (1 ≤ C ≤ 1000) which is the number of datasets that follow.

Each dataset consists of a single line of input containing a single integer N (1 ≤ N ≤ 1000), which is the size.

Output

For each dataset, there is to be one line of output consisting of: the dataset number starting at 1, a single space, the size, a single space and the number of visible points for that size.

Sample Input

4
2
4
5
231

Sample Output

1 2 5
2 4 13
3 5 21
4 231 32549

显然易得，除(1,0)(0,1)(1,1)三个钉子外，一个钉子能被看见当且仅当1<=x,y<=N且gcd(x,y)=1，对于每个y，这样的x数量恰好是φ(y)，本题答案就是3+2*∑φ(i)
可以利用线性筛的思想，在O(N)的时间内求出φ值

#include<cstdio>
const int N=1e3+10;
int phi[N],prime[N],p,n,c,v[N],sum[N],ca;
void euler()
{
	for(int i=2;i<=1000;i++)
	{
		if(!v[i])
		{
			v[i]=i;prime[p++]=i;
			phi[i]=i-1;
		}
		for(int j=0;j<p&&i*prime[j]<=1000;j++)
		{
			v[prime[j]*i]=prime[j];
			phi[i*prime[j]]=phi[i]*(i%prime[j]?prime[j]-1:prime[j]);
		} 
	}
}
int main()
{
	//freopen("in.txt","r",stdin);
    scanf("%d",&c);
    euler();
    for(int i=2;i<=1000;i++)sum[i]=sum[i-1]+phi[i];
    while(c--)
    {
    	scanf("%d",&n);
    	printf("%d %d %d\n",++ca,n,3+2*sum[n]);
	}
	return 0;
}