Poj 3090 visible lattice points Euler's Function

Link: http://poj.org/problem? Id = 3090

In the coordinate system, select a point from the point 0 ≤ x, y ≤ n in the x and y ≤ n, and these points do not pass through other points.

Idea: Obviously, the intersection of (0, 0) and (x, y) does not pass through other points only when x and y have mutual quality. For X, when Y is equal to N, all the points that can be selected are less than or equal to N and the number of mutual quality with N. A total of Euler (n) points are not overlapped. So we can obtain the recursive formula AA [I] = AA [I] + 2 * Euler (n ).
Code:

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <map>#include <cstdlib>#include <queue>#include <stack>#include <vector>#include <ctype.h>#include <algorithm>#include <string>#include <set>#define PI acos(-1.0)#define maxn 10005#define INF 0x7fffffff#define eps 1e-8typedef long long LL;typedef unsigned long long ULL;using namespace std;int aa[1005];int Euler(int tot){    int num=tot;    for(int i=2; i<=tot; i++)    {        if(tot%i==0)            num=num/i*(i-1);        while(tot%i==0)            tot/=i;    }    return num;}void init(){    aa[0]=0;    aa[1]=3;    for(int i=2; i<=1000; i++)        aa[i]=aa[i-1]+Euler(i)*2;}int main(){    int T;    scanf("%d",&T);    init();    for(int ii=1; ii<=T; ii++)    {        int tot;        scanf("%d",&tot);        printf("%d %d %d\n",ii,tot,aa[tot]);    }    return 0;}