POJ2478 Farey Sequence

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

 

欧拉函数模板题，16ms的O(n)线性筛，代码是抄贾教的。。

Codes:

#include<cstdio>
#include<iostream>
using namespace std;
int n,phi[1001000],tot,prime[500000];
long long sum[1001000];
bool check[1000100];
void PHI(int n){
    phi[1] = 1;
    for(int i=2;i<=n;i++){
        if(!check[i]){
            prime[++tot] = i;
            phi[i] = i - 1;
        }
        for(int j=1;j<=tot;j++){
            if(prime[j]*i>n) break;
            check[prime[j]*i] = true;
            if(i%prime[j]==0){
                phi[i*prime[j]] = phi[i] * prime[j];
                break;
            }else   phi[i*prime[j]] = phi[i] * (prime[j]-1);
        }
    }
}

int main(){
    PHI(1000000);
    sum[2] = 1;
    for(int i=3;i<=1000000;i++) sum[i]+=sum[i-1] + phi[i];
    while(scanf("%d",&n)!=EOF && n) cout<<sum[n]<<endl;
    return 0;
}