POJ2478(欧拉函数)

Farey Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15242		Accepted: 6054

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
思路：欧拉函数打表。

#include <cstdio>
using namespace std;
const int MAXN=1000005;
long long euler[MAXN];
void sieve()
{
    for(int i=1;i<MAXN;i++)    euler[i]=i;
    for(int i=2;i<MAXN;i+=2) euler[i]/=2;
    for(int i=3;i<MAXN;i+=2)
    {
        if(euler[i]==i)
        {
            for(int j=i;j<MAXN;j+=i)
            {
                euler[j]=euler[j]*(i-1)/i;
            }
        }
    }
    for(int i=3;i<MAXN;i++)
    {
        euler[i]+=euler[i-1];
    }
}
int main()
{
    sieve();
    int n;
    while(scanf("%d",&n)!=EOF&&n!=0)
    {
        printf("%lld
",euler[n]);
    }
    return 0;
}