POJ_2478_Farey Sequence

Farey Sequence

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 11067 Accepted: 4267

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

23450

Sample Output

1359

Source

POJ Contest,Author:Mathematica@ZSU

题型：数论

题意：求n以内所有互质的数的对数

分析：

       先预处理出10^7以内的所有的欧拉函数值φ( i )，然后累加从2到n的phi即可。（题目中将讲n<10^6，开了123456返了个RE，改成1234567就AC了，无解。。。）

代码：

#include<iostream>#include<cstring>#include<cstdio>#include<cmath>#define LL long longusing namespace std;const int maxn=1234567;int phi[maxn];void getphi(){    memset(phi,0,sizeof(phi));    phi[1]=1;    for(int i=2;i<maxn;i++){        if(!phi[i]){            for(int j=i;j<maxn;j+=i){                if(!phi[j])                    phi[j]=j;                phi[j]=phi[j]/i*(i-1);            }        }    }}int main(){    getphi();    int n;    while(scanf("%d",&n)&&n){        LL sum=0;        for(int i=2;i<=n;i++){            sum+=(LL)phi[i];        }        printf("%I64d\n",sum);    }    return 0;}