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

Submit Status Practice POJ 2478

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+5;
ll a[maxn];
void get()
{
    for(int i=2;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-1);
            }
        }
        a[i]+=a[i-1];
    }
}
int main()
{
    get();
    int n;
    while(cin>>n&&n)
    cout<<a[n]<<endl;
    return 0;
}