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

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
-----------------------------------------------------------------

和[SDOI2008]儀仗隊沒有什麽區別，只是加了個多數據。

[SDOI2008]儀仗隊 -》 http://www.cnblogs.com/noblex/p/7535245.html
還是放一下代碼吧

 1 #include <cstdio>
 2 #include <cmath>
 3 const int maxn=1005;
 4 int phi[maxn];
 5 int phis(int n)
 6 {
 7     for(int i=2;i<=n;i++) phi[i]=0;
 8     for(int i=2;i<=n;i++)
 9     {
10         if(!phi[i])
11         for(int j=i;j<=n;j+=i)
12         {
13             if(!phi[j]) phi[j]=j;
14             phi[j]=phi[j]/i*(i-1);
15         }
16     }
17     return 0;
18 }
19 int main()
20 {
21     int t,n;
22     scanf("%d",&t);
23     phis(maxn);
24     for(int i=1;i<=t;i++)
25     {
26         long long sum=0;
27         scanf("%d",&n);
28         for(int j=2;j<=n;j++) sum+=phi[j];
29         sum*=2;sum+=3;
30         printf("%d %d %lld\n",i,n,sum);
31     }
32     return 0;
33 }

【POJ】3090 Visible Lattice Points