POJ 3090 Visible Lattice Points

Visible Lattice Points

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6662 Accepted: 3984

Description

A lattice point (x, y) in the first quadrant (x andy 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

4245231

Sample Output

1 2 52 4 133 5 214 231 32549

分析：可以dp思想，每一个n都可以通过n-1得到，只要计算最外层的两条边上的点是不是互质即可。

代码：

#include<iostream>#include<cstdio>#include<cstring>using namespace std;long long num[1005];int c, n, cas = 1;int gcd(int a, int b){return b == 0 ? a : gcd(b, a%b);}void f(){memset(num, 0, sizeof(num));num[1] = 3;for (int idx = 2; idx <= 1000; ++idx)for (int x = 1; x < idx; ++x)if (gcd(x, idx) == 1)num[idx] += 2;for (int i = 2; i <= 1000; ++i)num[i] += num[i - 1];}int main(){f();scanf("%d", &c);while (c--){scanf("%d", &n);printf("%d %d %lld\n", cas++, n, num[n]);}}