poj3090(eular)

http://poj.org/problem?id=3090

Visible Lattice Points

Time Limit: 1000MSMemory Limit: 65536KTotal Submissions: 4455Accepted: 2535

Description

A lattice point (x,y) in the first quadrant (x andy are integers greater than or equal to 0), otherthan the origin, is visible from the origin if the line from (0, 0)to (x, y) does not pass through anyother lattice point. For example, the point (4, 2) is not visiblesince the line from the origin passes through (2, 1). The figurebelow shows the points (x, y) with 0 ≤x, y ≤ 5 with lines from the origin tothe 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 singleinteger C (1 ≤ C ≤ 1000) which is thenumber of datasets that follow.

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

Output

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

Sample Input

4245231

Sample Output

1 2 52 4 133 5 214 231 32549

Source

Greater New York 2006

题意：要求坐标系第一象限中从原点能看到的点数。图会是沿着k=1斜率对称的，对上半部分做分析，x值做分子，y值做分母，会发现x，y必须互为质数才能满足，因为如果是

4/6,肯定会然2/3给覆盖了

所以得找出小于n与n互为质数的个数（欧拉函数），然后乘2就行。

牛人做法：

先画一条(0, 0)到(n, n)的线，把图分成两部分，两部分是对称的，只需算一部分就好。取右下半，这一半里的点(x, y)满足x >= y可以通过欧拉函数计算第k列有多少点能够连到(0, 0)若x与k的最大公约数d > 1，则(0, 0)与(x, k)点的连线必定会通过(x/d, k/d)，就被挡住了所以能连的线的数目就是比k小的、和k互质的数的个数，然后就是欧拉函数。

一开始想找规律的发现无规律可循，以为自己找的规律错了。

#include 
#include
#include
using namespace std;
int eular(int n)//
{
    int ret=1,i;
    for(i=2;i*i<=n;i++)
    {
        if(n%i==0)
        {
            n/=i,ret*=i-1;
            while(n%i==0)
            {
                n/=i,ret*=i;
            }
        }
    }
    if(n>1)
    {
        ret*=n-1;
    }
    return ret;
}
int main()
{
    int t;
    scanf("%d",&t);
    int cases=1;
    while(t--)
    {
        int n;
        scanf("%d",&n);
        int sum=0;
        for(int i=2;i<=n;i++)
        {
            sum+=eular(i);
        }
        printf("%d %d %d\n",cases++,n,sum*2+3);
    }
    return 0;
}