【POJ】3090 Visible Lattice Points（欧拉函数）

最新推荐文章于 2021-11-06 13:58:01 发布

转载最新推荐文章于 2021-11-06 13:58:01 发布 · 50 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/noblex/p/7535258.html

本文介绍了一种计算在第一象限中从原点可见的格点数量的方法，并提供了一个具体的算法实现。可见的格点是指从原点到该点的直线不经过其他整数坐标点。文章还附带了一个C++代码示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Visible Lattice Points

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7705		Accepted: 4707

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.

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

Sample Output

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

分析：和[SDOI2008]仪仗队没有什么区别，只是加了个多数据。

[SDOI2008]仪仗队 -》 http://www.cnblogs.com/noblex/p/7535245.html
还是放一下代码吧

 1 #include <cstdio>
 2 #include <cmath>
 3 const int maxn=1005;
 4 int phi[maxn];
 5 int phis(int n)
 6 {
 7     for(int i=2;i<=n;i++) phi[i]=0;
 8     for(int i=2;i<=n;i++)
 9     {
10         if(!phi[i])
11         for(int j=i;j<=n;j+=i)
12         {
13             if(!phi[j]) phi[j]=j;
14             phi[j]=phi[j]/i*(i-1);
15         }
16     }
17     return 0;
18 }
19 int main()
20 {
21     int t,n;
22     scanf("%d",&t);
23     phis(maxn);
24     for(int i=1;i<=t;i++)
25     {
26         long long sum=0;
27         scanf("%d",&n);
28         for(int j=2;j<=n;j++) sum+=phi[j];
29         sum*=2;sum+=3;
30         printf("%d %d %lld\n",i,n,sum);
31     }
32     return 0;
33 }