POJ-3090 Visible Lattice Points(欧拉函数)

Visible Lattice Points

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6826		Accepted: 4103

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 2006

 

其实吧，和前前道题是一模一样的，然而我这个傻逼写了三次欧拉函数了，还是不能一遍写对QAQ

#include <cstdio>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <queue>
#include <stack>
#include <vector>
#include <iostream>
#include "algorithm"
using namespace std;
typedef long long LL;
const int MAX=1005;
int n,T;
int phi[MAX];
void euler(){
    int i,j;
    for (i=1;i<MAX;i++) phi[i]=i;
    for (i=2;i<MAX;i+=2) phi[i]/=2;
    for (i=3;i<MAX;i+=2)
     if (phi[i]==i)
      for (j=i;j<MAX;j+=i)
       phi[j]=phi[j]/i*(i-1);
}
int main(){
    freopen ("visible.in","r",stdin);
    freopen ("visible.out","w",stdout);
    int i,j,cas(0);
    scanf("%d",&T);
    euler();
    while (T--){
        LL ans(0);
        scanf("%d",&n);
        for (i=1;i<=n;i++)
         ans=ans+(LL)phi[i];
        printf("%d %d %lld
",++cas,n,ans*2+1);
    }
    return 0;
}