POJ - 2478 Farey Sequence

Farey Sequence

Time Limit: 1000MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

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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 <= 10  ⁶). 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

欧拉函数的简单题，跟poj3090差不多。

代码：

#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=1000005;
int phi[maxn];
long long  s[maxn];//s[]用long long
void getphi(int n)//欧拉函数打表
{
    int i,j;
    memset(phi,0,sizeof(phi));
    phi[1]=1;
    for(i=2;i<=n;i++)
    {
        if(!phi[i])
        for(j=i;j<=n;j+=i)
        {
            if(!phi[j])
            phi[j]=j;
            phi[j]=phi[j]/i*(i-1);
        }
    }
}
int main()
{
    int n,i;
    getphi(maxn);
    for(i=2;i<maxn;i++)
    s[i]=phi[i]+s[i-1];//把s[]打表，降低时间复杂度
    while(scanf("%d",&n)&&n)
        printf("%lld\n",s[n]);
    return 0;
}