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.
题意：求从源点看到的点的个数。斜率可约分的都可以被阻挡，每次添加的符合条件的个数就是分子和分母互质的个数。即为对于一个数n，求小于n的与n互质的数的个数。
即为欧拉函数：
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;
}
AC代码：
#include <cstdio>
#include <iostream>
using namespace std;
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;
}
int solve(int n)//递归求点数
{
int sum = 0;
if (n == 1)//如果是1X1的话，只有一个点
return 3;
sum = solve(n - 1);
sum += 2 * Euler(n);
return sum;
}
int main()
{
int T;
scanf("%d", &T);
for (int i = 1; i <= T; i++)
{
int n;
scanf("%d", &n);
printf("%d %d %d\n", i, n, solve(n));
}
return 0;
}