POJ 2407 Relatives（欧拉函数入门题）

最新推荐文章于 2020-07-17 07:59:37 发布

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原文链接：http://www.cnblogs.com/Lis-/p/9387619.html

本文解析了一道典型的欧拉函数模板题，介绍了如何通过欧拉函数计算与n互质的正整数数量。该文提供了完整的C++代码实现，并采用O(根号n)的复杂度进行求解。

Relatives

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

题意：输入个n，求（1，n-1）与n互质的个数有多少个
思路：这明显就是欧拉函数模版题，就是求n的欧拉数

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long ll;
ll phi(ll n)//求n的欧拉数
{
    ll sum=n;
    for(int i=2;i*i<=n;i++)//O（根号n的复杂度）
    {
        if(n%i==0)
        {
            sum=sum-sum/i;
            do
            {
                n/=i;
            }while(n%i==0);
        }
    }
    if(n>1) sum=sum-sum/n;
    return sum;
}
int main()
{
    ll n;
    while(scanf("%lld",&n)!=EOF)
    {
        if(n==0)
        break;
        printf("%lld\n",phi(n));
    }
}