poj 2478 Farey Sequence

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本文介绍了一种利用欧拉函数和前缀和的方法来高效计算Farey序列中项的数量。通过具体实例展示了如何计算特定整数n内的互质数对数量。

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Farey Sequence

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 15288 Accepted: 6070

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

【分析】
题意：求n以内有多少对互质的数
欧拉函数+前缀和。
刷yhx的水题真是令人愉悦…

【代码】

//poj 2478 Farey Sequence
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#define LL long long
#define fo(i,j,k) for(i=j;i<=k;i++)
using namespace std;
const int n=1e6;
int phi[n+5],p[n+5],k;
bool vis[n+5];
LL ans[n+5];
void get()
{
    int i,j;
    fo(i,2,n)
    {
        if(!vis[i]) p[++p[0]]=i,phi[i]=i-1;
        for(j=1;j<=p[0] && i*p[j]<=n;j++)
        {
            vis[i*p[j]]=1;
            if(i%p[j]==0)
            {
                phi[i*p[j]]=phi[i]*p[j];
                break;
            }
            else phi[i*p[j]]=phi[i]*(p[j]-1);
        }
        ans[i]=ans[i-1]+phi[i];
    }
}
int main()
{
    get();
    while(scanf("%d",&k) && k)
      printf("%lld\n",ans[k]);
    return 0;
}