（x, y）被看到僅當x與y互質

由此聯想到尤拉函式

x=y是1個點，然後把正方形分成兩半，一邊是φ（n）

所以答案是2*φ（n）+1

#include<cstdio>
#include<cctype>
#define REP(i, a, b) for(int i = (a); i < (b); i++) 
#define _for(i, a, b) for(int i = (a); i <= (b); i++) 
using namespace std;

typedef long long ll;
const int MAXN = 1123;
ll euler[MAXN];

void get_euler()
{
	_for(i, 1, MAXN) euler[i] = i;
	_for(i, 2, MAXN)
	{
		if(euler[i] == i)
			for(int j = i; j <= MAXN; j += i)
				euler[j] = euler[j] / i * (i - 1);
		euler[i] += euler[i-1]; 
	}
}

void read(ll& x)
{
	int f = 1; x = 0; char ch = getchar();
	while(!isdigit(ch)) { if(ch == '-1') f = -1; ch = getchar(); }
	while(isdigit(ch)) { x = x * 10 + ch - '0'; ch = getchar(); }
	x *= f;
}

int main()
{
	get_euler();
	ll n; read(n);
	_for(i, 1, n)
	{
		ll x; read(x);
		printf("%d %lld %lld\n", i, x, 2 * euler[x] + 1);
	}
	return 0;
}