POJ 2478 Farey Sequence(欧拉函数前n项和)

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

题意:求欧拉函数的前n项和。

题解:打表筛选即可。

#include <iostream>
using namespace std;
typedef long long ll;
const int maxn=1e6+;
ll a[maxn];
void get()
{
for(int i=;i<maxn;i++)
{
if(!a[i])
{
for(int j=i;j<maxn;j+=i) //已经包含了素筛的思想
{
if(!a[j])
a[j]=j;
a[j]=a[j]/i*(i-);
}
}
a[i]+=a[i-];
}
}
int main()
{
get();
int n;
while(cin>>n&&n)
cout<<a[n]<<endl;
return ;
}
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