pku3090 Visible Lattice Points：人生第一個歐拉函數

pku3090 Visible Lattice Points：人生第一個歐拉函數

這是一道很有紀念價值的題目！

 

Visible Lattice Points Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5812   Accepted: 3434

 

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.

Sample Input

4
2
4
5
231

Sample Output

1 2 5
2 4 13
3 5 21
4 231 32549

Source

Greater New York 2006 既然這是一道有價值的題目，我就不給翻譯了

直接上代碼！

 

#include 
#include 
#include 
#include 
#define __________ 1100
#define ___________ 0
#define ____________ 1
#define _____________ 2
#define ______________ 3
using 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 ___________;
}

人生第一個歐拉函數！代碼極其精致！各位看了記得給我點贊！我的代碼風格可是我們小學最好的！

 

我也不多說什麼，自己看代碼吧！