POJ2478 Farey Sequence

 

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15023		Accepted: 5962

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

 

简单分析一波就知道，读入n时输出1~n的欧拉函数和即可。

飞快地敲了个暴力欧拉函数交上去，TLE。

默默打了欧拉函数表，WA。

然后把int换成long long，终于过了。

/*by SilverN*/
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
long long f[1000002];
int n;
void phi(){
    int i,j;
    for(i=2;i<=1000000;i++)
        if(!f[i])
            for(j=i;j<=1000000;j+=i){
                if(!f[j])f[j]=j;
                f[j]=f[j]/i*(i-1);
            }
}
int main(){
    phi();
    for(int i=1;i<=1000000;i++){
        f[i]+=f[i-1];//求前缀和 
    }
    while(scanf("%d",&n) && n){
        cout<<f[n]<<endl;
    }
    return 0;
}