Problem J:Relatives（欧拉函数）

最新推荐文章于 2018-08-23 11:02:00 发布

sdau20171989

最新推荐文章于 2018-08-23 11:02:00 发布

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分类专栏：数学——数论例题

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本文介绍了一种使用欧拉函数计算小于给定整数n的所有与n互素的正整数数量的方法，并提供了一个具体的C++实现案例。通过预先计算欧拉函数值，快速解决特定范围内的互素数对计数问题。

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http://poj.org/problem?id=2407

求n以内互素的无序数对数（欧拉函数打表）

Description

Given n, a positive integer, how many positive integers less than n are relatively prime to n? Two integers a and b are relatively prime if there are no integers x > 1, y > 0, z > 0 such that a = xy and b = xz.

Input

There are several test cases. For each test case, standard input contains a line with n <= 1,000,000,000. A line containing 0 follows the last case.

Output

For each test case there should be single line of output answering the question posed above.

Sample Input

7
12
0

Sample Output

6
4

Source Code:

#include<iostream>
#define N 1000010
#define ll long long
using namespace std;
ll phi[N],ans[N];
ll n;
void Euler()//打表N以内数的欧拉函数
{
    for(ll i=1;i<=N;i++)
        phi[i]=i;
    for(ll i=2;i<=N;i+=2)
        phi[i]/=2;
    for(ll i=3;i<=N;i+=2)
    {
        if(phi[i]==i)
        {
            for(ll j=i;j<=N;j+=i)
                phi[j]=phi[j]/i*(i-1);
        }
    }
}
int main()
{
    Euler();
    while(cin>>n&&n)
    {
        for(ll i=2;i<=n;i++)
            ans[i]=ans[i-1]+phi[i];
        cout<<ans[n]<<endl;
    }
    return 0;
}