poj 2478: Farey Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13984		Accepted: 5526

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

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<queue>
#include<vector>
using namespace std;
typedef long long LL;
const LL maxn=1e6+5;
LL N,tot,ANS;
LL prime[maxn],phi[maxn];
bool not_p[maxn];
void shai(){
    not_p[1]=1;
    for(LL i=1;i<maxn;i++){
        if(not_p[i]==false){
            prime[++tot]=i;
            phi[i]=i-1;
        }
        for(LL j=1;j<=tot;j++){
            LL k=prime[j]*i;
            if(k>maxn) break;
            not_p[k]=true;
            if(i%prime[j]!=0){
                phi[k]=phi[i]*phi[prime[j]];
            }
            else{
                phi[k]=phi[i]*prime[j];
                break;
            }
        }
    }
}
int main(){
    shai();
    while(scanf("%lld",&N)&&N){
        ANS=0;
        for(LL i=2;i<=N;i++){
            ANS+=phi[i];
        }
        printf("%lld\n",ANS);
    }
    return 0;
}