 A - Farey Sequence——（筛法求欧拉函数）

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A - Farey Sequence——（筛法求欧拉函数）

angel_imp 2016-03-22 17:30:00 浏览822

A - Farey Sequence
Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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 <= 10 6). 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

2
3
4
5
0

Sample Output

1
3
5
9

void Init()
{
memset(phi, 0, sizeof(phi)) ;
for(int i=2; i<MAXN; i++)
{
if(!phi[i])
{
for(int j=i; j<MAXN;j+=i )
{
if(!phi[j])
phi[j] = j ;
phi[j] = phi[j]/i*(i-1) ;
}
}
}
}

Code：

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long LL;
const int MAXN = 1e6+5;

int phi[MAXN];
void Init()
{
memset(phi, 0, sizeof(phi)) ;
for(int i=2; i<MAXN; i++)
{
if(!phi[i])
{
for(int j=i; j<MAXN;j+=i )
{
if(!phi[j])
phi[j] = j ;
phi[j] = phi[j]/i*(i-1) ;
}
}
}
}
int main()
{
Init();
int m;
while(cin>>m,m)
{
LL sum = 0;
for(int i=2; i<=m; i++)
sum += phi[i];
cout<<sum<<endl;
}
return 0;
}

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