【poj 2478】 Farey Sequence

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

这道题是一个典型的欧拉函数，先预处理出1-1000000之间所有数的欧拉函数值，再求一边和数组就可以了，下面是程序：

#include<stdio.h>
#include<iostream>
#define ll long long
using namespace std;
const int N=1000005;
bool p[N];
ll phi[N];
int read(){
	char c=getchar();
	int s=0;
	while(c<'0'||c>'9'){
		c=getchar();
	}
	while(c>='0'&&c<='9'){
		s*=10;
		s+=c-'0';
		c=getchar();
	}
	return s;
}
void work(){
	p[0]=p[1]=1;
	int i,j;
	for(i=2;i<=1000000;i++){
		phi[i]=i;
	}
	for(i=2;i*i<=1000000;i++){
		if(!p[i]){
			for(j=i*i;j<=1000000;j+=i){
				p[j]=1;
			}
		}
	}
	for(i=2;i<=1000000;i++){
		if(!p[i]){
			for(j=i;j<=1000000;j+=i){
				phi[j]=phi[j]/i*(i-1);
			}
		}
	}
	for(i=2;i<=1000000;i++){
		phi[i]+=phi[i-1];
	}
}
void out(ll x){
	if(x>9){
		out(x/10);
	}
	putchar(x%10+'0');
}
int main(){
	work();
	int n;
	while(n=read()){
		out(phi[n]);
		putchar('\n');
	}
	return 0;
}