POJ 2407：Relatives（欧拉函数模板）

Relatives

AC代码

                                             Relatives

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16186		Accepted: 8208

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

欧拉函数模板题

AC代码

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <math.h>
#include <limits.h>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <set>
#include <string>
#define ll long long
#define ms(a) memset(a,0,sizeof(a))
#define pi acos(-1.0)
#define INF 0x3f3f3f3f
const double E=exp(1);
using namespace std;
const int maxn=1e6+10;
int a[maxn];
int b[maxn];
//欧拉函数公式：
//φ(N)=N*(1-1/P1)*(1-1/P2)*...*(1-1/Pn).
int main(int argc, char const *argv[])
{
	int n;
	while(cin>>n&&n)
	{
		ms(b);
		ll ans=n;
		int vis=n;
		for(int i=2;i*i<=vis;i++)
		{
			if(vis%i==0)
			{
				ans=ans/i*(i-1);
				while(vis%i==0)
				{
					vis/=i;
				}	
			}
		}
		if(vis>1)
			ans=ans/vis*(vis-1);
		cout<<ans<<endl;
	}
	return 0;
}

// 两种求欧拉函数的方法：
// 
// ******************第一种直接求解****************
// long long Euler(long long x)
// {
//     int res = x,a = x;
//     for(int i=2;i*i<=a;i++)
//     {
//         if(a%i==0)
//         {
//             res = res/i*(i-1);//res -= res/i;
//             while(a%i==0)a/=i;
//         }
//     }
//     if(a>1)res =res/a*(a-1);//res -= res/a;
//     return res;
// }
 // ******************第二种打表求解****************
// #define Max 1000001
// int euler[Max];
// void Euler()
// {
//      euler[1]=1;
//      for(int i=2;i<Max;i++)
//        euler[i]=i;
//      for(int i=2;i<Max;i++)
//         if(euler[i]==i)
//            for(int j=i;j<Max;j+=i)
//               euler[j]=euler[j]/i*(i-1);//先进行除法是为了防止中间数据的溢出
// }