poj 2478 Farey Sequence 欧拉函数前缀和

Farey Sequence

Time Limit: 1000MS		Memory Limit: 65536K

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

2
3
4
5
0

Sample Output

1
3
5
9

Source

POJ Contest,Author:Mathematica@ZSU

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
#define ll long long
#define esp 1e-13
const int N=1e3+10,M=1e6+1000,inf=1e9+10,mod=1000000007;
ll p[M],ji;
bool vis[M];
ll phi[M];
ll sum[M];
void get_eular(int n)
{
    ji = 0;
    memset(vis, true, sizeof(vis));
    for(int i = 2; i <= n; i++)
    {
        if(vis[i])
        {
            p[ji ++] = i;
            phi[i] = i - 1;
        }
        for(int j = 0; j < ji && i * p[j] <= n; j++)
        {
            vis[i * p[j]] = false;
            if(i % p[j] == 0)
            {
                phi[i * p[j]] = phi[i] * p[j];
                break;
            }
            else
            phi[i * p[j]] = phi[i] * phi[p[j]];
        }
    }
}
int main()
{
    int x,i;
    get_eular(M);
    memset(sum,0,sizeof(sum));
    for(i=1;i<=1e6;i++)
    sum[i]=sum[i-1]+phi[i];
    while(~scanf("%d",&x))
    {
        if(!x)break;
        printf("%lld
",sum[x]);
    }
    return 0;
}