这是一道很有纪念价值的题目！
Visible Lattice Points
Time Limit:1000MSMemory Limit:65536K
Total Submissions:5812Accepted:3434
Description
A lattice point (x,y) in the first quadrant (xandyare 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 integerC(1 ≤C≤ 1000) which is the number of datasets that follow.
Each dataset consists of a single line of input containing a single integerN(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
Source
既然这是一道有价值的题目，，我就不给翻译了
直接上代码！
#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>#define __________ 1100#define ___________ 0#define ____________ 1#define _____________ 2#define ______________ 3using namespace std;int __[__________], _[__________], _______, ____;bool ___[__________];void getprime (){memset ( ___, false, sizeof (___) );memset ( _, ___________, sizeof (_) );____ = ___________;int _____, ______;for ( _____ = _____________; _____ <= _______; _____ ++ ){if ( ___[_____] == false ){__[++____] = _____;_[_____] = _____-____________;}for ( ______ = ____________; ______ <= ____ && _____*__[______] <= _______; ______ ++ ){___[_____*__[______]] = true;if ( _____ % __[______] == ___________ ){_[_____*__[______]] = _[_____] * __[______];}else {_[_____*__[______]] = _[_____] * ( __[______] – ____________ );}}}for ( _____ = ____________; _____ <= _______; _____ ++ ) _[_____] += _[_____-____________];}int main (){int _____, ______, ________, _________;_______ = __________;getprime ();scanf ( "%d", &________ ); _________ = ___________;while ( ________ — ){scanf ( "%d", &_______ );printf ( "%d %d %d\n", ++_________, _______, _[_______]*_____________+______________ );}return ___________;}人生第一个欧拉函数！代码极其精致！各位看了记得给我点赞！我的代码风格可是我们小学最好的！
我也不多说什么，自己看代码吧！
上帝助自助者。
