POJ 3090 Visible Lattice Points

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

其实就是求欧拉函数，数据小暴力求也行。

#include<set>
#include<map>
#include<ctime>
#include<cmath>
#include<stack>
#include<queue>
#include<bitset>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define rep(i,j,k) for (int i = j; i <= k; i++)
#define per(i,j,k) for (int i = j; i >= k; i--)
#define loop(i,j,k) for (int i = j;i != -1; i = k[i])
#define lson x << 1, l, mid
#define rson x << 1 | 1, mid + 1, r
#define ff first
#define ss second
#define mp(i,j) make_pair(i,j)
#define pb push_back
#define pii pair<int,LL>
#define in(x) scanf("%d", &x);
using namespace std;
typedef long long LL;
const int low(int x) { return x&-x; }
const double eps = 1e-4;
const int INF = 0x7FFFFFFF;
const int mod = 998244353;
const int N = 1e6 + 10;
int n, sum[N];
int phi[N], f[N], p[N], t, T;

void init()
{
  t = 0;  sum[1] = phi[1] = f[1] = 1;
  rep(i, 2, N - 1)
  {
    if (!f[i]) p[t++] = i, phi[i] = i - 1;
    for (int j = 0,k; j < t&&p[j] * i < N; j++)
    {
      k = p[j] * i; f[k] = 1;
      phi[k] = phi[i] * (p[j] - 1);
      if (i % p[j] == 0) { phi[k] = phi[i] * p[j]; break; }
    }
    sum[i] = sum[i - 1] + phi[i] * 2;
  }
}

int main()
{
  init(); in(T);
  int cas = 0;
  while (T--)
  {
    scanf("%d", &n);
    printf("%d %d %d\n", ++cas, n, sum[n] + 2);
  }
  return 0;
}