POJ 2478Farey Sequence

2018-01-27 11:14:37来源:cnblogs.com作者:自为风月马前卒人点击

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Farey Sequence
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 17744 Accepted: 7109

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  std的玄学做法没看懂
给n,求ans[n]。其中$ans[n]=ans[n-1]+phi[n]$,且n的范围比较大,在10的6次以内。则考虑打表解决。 
先得到能整除i的最小正整数$md[i]$(一定是个素数),再利用性质3,得到$phi[i]$
不过我用线性筛水过去啦。 
#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#define LL long long using namespace std;const LL MAXN=3*1e6+10;LL prime[MAXN],tot=0,vis[MAXN],phi[MAXN],N;void GetPhi(){    for(LL i=2;i<=N;i++)    {        if(!vis[i])        {            prime[++tot]=i;            phi[i]=i-1;        }        for(LL j=1;j<=tot&&prime[j]*i<=N;j++)        {            vis[ i*prime[j] ] = 1;             if(i%prime[j]==0)            {                phi[ i*prime[j] ]=phi[i]*prime[j];                break;            }            else phi[ i*prime[j] ]=phi[i]*(prime[j]-1);        }    }    for(LL i=1;i<=N;i++)        phi[i]=phi[i]+phi[i-1];}int main(){    N=2*1e6+10;    GetPhi();    while(cin>>N&&N!=0)        printf("%lld/n",phi[N]);    return 0;}

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