欧拉定理 A - Relatives （模板）

最新推荐文章于 2019-08-08 14:40:00 发布

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最新推荐文章于 2019-08-08 14:40:00 发布

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本文介绍了一个计算欧拉函数的C++程序，该程序用于找出小于给定正整数且与其互质的所有正整数的数量。通过分解给定整数的素因子并应用欧拉函数的公式来解决这一数学问题。

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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

这道题就是求比一个数小又和这个数互质的个数，就是欧拉值，这是前人已经推导出来了；

phi=n/(P1)*(P1-1)/P2*(P2-1)....../(Pn-1)*(Pn-1 - 1)/Pn*(Pn - 1),带公式就行；P是素因子

#include<iostream>
#include<math.h>
#include<string.h>
#include<cstdio>
#define maxn 100000
#define ll long long
using namespace std;
int ans;
int solve(int x)
{
	ans=x;
	for(int i=2;i*i<x;i++)
	{
		if(x%i==0)
		ans=ans/i*(i-1);
		while(x%i==0) x/=i;
	}
	if(x>1)
	ans=ans/x*(x-1);
}
int main()
{
	ll n;
	while(scanf("%lld",&n),n)
	{
		solve(n);
		cout<<ans<<endl;
	}
	return 0;
}