Visible Lattice Points（POJ3090）

【题目描述】
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.
【输入格式】
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.
【输出格式】
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.
【样例输入】
4
2
4
5
231
【输出格式】
1 2 5
2 4 13
3 5 21
4 231 32549
【分析】
不难发现一个规律：对于n*n，可以从(0,0)连接到(n,0)到(n,n)上，斜率将会是1/n,2/n,…,(n-1)/n。
凡是分子和分母能够约分的，也就是有公约数，前面都已经有过了，所以每次添加的个数就是分子和分母互质的个数。
于是问题变成：对于正整数n，求小于n且与n互质的数的个数。这就是欧拉函数模板题了。

#include<cstdio>const int Maxn=1000;bool vis[Maxn+5];int phi[Maxn+5],p[Maxn+5];int n,ans,t;int main(){    phi[1]=1;    for (int i=2;i<=Maxn;i++){        if (!vis[i]) p[++p[0]]=i,phi[i]=i-1;        for (int j=1;j<=p[0] && i*p[j]<=Maxn;j++){            vis[i*p[j]]=1;            if (i%p[j]==0){ phi[i*p[j]]=phi[i]*p[j];break; } else phi[i*p[j]]=phi[i]*(p[j]-1);        }        ans+=phi[i];    }    scanf("%d",&t);    int Case=0;    while (t--){        scanf("%d",&n);        ans=0;        for (int i=1;i<=n;i++) ans+=phi[i];        printf("%d %d %d\n",++Case,n,ans*2+1);    }    return 0;}