poj2478 又一歐拉公式的運用

Farey Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7881		Accepted: 2957

Description

The Farey Sequence Fn for any integer n with n >= 2 is the set of irreducible rational numbers a/b with 0 < a < b <= n and gcd(a,b) = 1 arranged in increasing order. The first few are 
F2 = {1/2} 
F3 = {1/3, 1/2, 2/3} 
F4 = {1/4, 1/3, 1/2, 2/3, 3/4} 
F5 = {1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5} 

You task is to calculate the number of terms in the Farey sequence Fn.

Input

There are several test cases. Each test case has only one line, which contains a positive integer n (2 <= n <= 106). There are no blank lines between cases. A line with a single 0 terminates the input.

Output

For each test case, you should output one line, which contains N(n) ---- the number of terms in the Farey sequence Fn. 

Sample Input

23450

Sample Output

1359

題目大意就是求Fi集合中元素的個數，其中Fi集合的元素滿足下列條件

形如a/b，且0<a<b<=i, gcd(a,b)=1 

很明顯，這題就是歐拉公式的運用，關於歐拉公式可查看下這篇文章

http://www.cnblogs.com/ACShiryu/archive/2011/08/04/poj2407.html

對於這題，可以先求出以每一個小於m的數為分母的數的個數，即也是與該數互素的數的個數，也就是求的phi[i]；

然後再每一個phi都加起來

題目就是比較簡單的歐拉運用，1A

參考代碼：

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int phi[1000005];
int main()
{
int i , j ;
    memset ( phi , 0 ,sizeof ( phi ) ) ;
for ( i = 2 ; i <= 1000000 ; i ++ )
    {//篩選求phi
        if ( ! phi [i] )
        {
for ( j = i ; j <= 1000000 ; j += i )
            {
if ( ! phi [j] )
                    phi [j ] = j ;
                phi [j] = phi [j] / i * ( i - 1 ) ;
            }
        }
    }
int n ;
while ( cin >> n  , n )
    {
        __int64 sum = 0 ;
for ( i = 2 ; i <= n ; i ++ )
            sum += phi [i] ;
        cout<<sum<<endl;
    }
return 0;
}