POJ3090_Visible Lattice Points【欧拉函数】

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




4
2
4
5
231
Sample Output




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

Greater New York 2006Visible Lattice Points

题目大意：如今有一个二维坐标系，仅仅有离散的整数坐标上有点。
如今站在N点向周围看去。问能看到多少个点。

假如看到了（2,1），那么（2,1）后边的（4,2）（6,3）…就被挡住
看不到了。
考虑1*1的时候，有三个点（1,0）（1,1）（0,1）。
（1,0）和（0,1）关于（1,1）对称
再看2*2的时候，有个点（1,0）（1,1）（2,1）（0,1）（1,2）
（1,0）和（0,1）关于（1,1）对称
（2,1）和（1,2）关于（1,1）对称
比1*1多了两个点。而且都是关于(1,1)对称，而（2,2）则被（1,1）遮挡住了
所以我们仅仅考虑下三角的情况。得出结果*2+1就是终于答案。
由于同斜率的点都被第一个点盖掉看不到了。所以我们仅仅考虑斜率有多少种就是得出结果了。
1*1的时候。斜率有0
2*2的时候，斜率有0,1/2
3*3的时候，斜率有0,1/2,1/3,2/3
4*4的时候，斜率有0,1/2(2/4),1/3,2/3,1/4,3/4;
5*5的时候，斜率有0,1/2(2/4),1/3,2/3,1/4,3/4,1/5,2/5,3/5,4/5
6*6的时候，斜率有0,1/2(2/4,3/6),1/3(2/6),2/3(4/6),1/4,3/4,1/5,2/5,3/5,4/5,1/6,5/6
能够看出，事实上就是求分母小于等于N的真分数有多少
那么就是单纯的欧拉函数了，这里用普通欧拉函数和高速求欧拉函数都能够

參考博文：http://blog.csdn.net/zhang20072844/article/details/8108727

#include<stdio.h>

int prime[1010],phi[1001];
bool unprime[1010];

void Euler()//高速求欧拉函数
{
    int i,j,k = 0;

    for(i = 2; i <= 1000; i++)
    {
        if(!unprime[i])
        {
            prime[k++] = i;
            phi[i] = i-1;
        }

        for(j = 0; j < k && i*prime[j] <= 1000; j++)
        {
            unprime[prime[j]*i] = true;
            if(i % prime[j] != 0)
                phi[prime[j]*i] = phi[i] * (prime[j]-1);
            else
            {
                phi[prime[j]*i] = phi[i] * prime[j];
                break;
            }
        }
    }
}
int main()
{
    int C,n;
    Euler();
    phi[1]=1;
    scanf("%d",&C);
    int kase = 1;
    while(C--)
    {
        scanf("%d",&n);
        int sum = 0;
        for(int i = 1;i <= n; i++)
            sum += phi[i];
        printf("%d %d %d\n",kase++,n,2*sum+1);
    }
    return 0;
}

#include <stdio.h>
#include <math.h>
int Euler(int n)//普通求欧拉函数
{
    int i,ret = n;
    for(i = 2; i <= sqrt(1.0*n); i++)
    {
        if(n % i == 0)
        {
            ret = ret - ret/i;
        }
        while(n % i == 0)
            n /= i;
    }
    if(n > 1)
        ret = ret - ret/n;
    return ret;
}
int main()
{

    int C,n;
    scanf("%d",&C);
    int kase = 1;
    while(C--)
    {
        scanf("%d",&n);
        int sum = 0;
        for(int i = 1;i <= n; i++)
            sum += Euler(i);
        printf("%d %d %d\n",kase++,n,2*sum+1);
    }
}