POJ 3090

Visible Lattice Points

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8397		Accepted: 5144

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

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刚开始脑子简单去暴力循环。。当然是T掉啦！

原来一直不明白欧拉函数，f(n)代表小于n的正整数中与n互质的数的数目。。就是这个题嘛。。记住了。

 

#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include<iostream>
#include <cmath>
#include<string>
#define ll long long 
#define dscan(a) scanf("%d",&a)
#define mem(a,b) memset(a,b,sizeof a)
using namespace std;
#define MAXL 1105
#define maxn 1000005
int f[MAXL];
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-') f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') {x=10*x+ch-'0';ch=getchar();}
    return x*f;
}
int p[MAXL], pNum, phi[MAXL];
void ola()
{
    int i, j;
    for(i = 2; i < MAXL; i++) {
        if(!f[i]) { p[pNum++] = i; phi[i] = i-1;}
        for(j = 0; j < pNum && p[j] * i < MAXL; j++ ) {
            f[p[j]*i] = 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);
        }
    }
}
int main()
{
    int n,c;
    cin>>n;int hh=1;
    ola();
    while(n--)
    {
        int c=read();
        phi[1]=1;
        cout<<hh++<<" "<<c<<" ";
        ll ans=0;
        for(int i=1;i<=c;++i) ans+=phi[i];
        cout<<ans*2+1<<endl;
    }
}