POJ - 2407 Relatives

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16494		Accepted: 8368

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

Waterloo local 2002.07.01

利用欧拉函数和它本身不同质因数的关系，用筛法计算出某个范围内所有数的欧拉函数值。

　　欧拉函数和它本身不同质因数的关系：欧拉函数ψ（Ｎ）＝Ｎ｛∏ｐ｜Ｎ｝（１－１/ｐ）。（Ｐ是数Ｎ的质因数）

　　如：

　　ψ（１０）＝１０×（１－１/２）×（１－１/５）＝４；

　　ψ（３０）＝３０×（１－１/２）×（１－１/３）×（１－１/５）＝８；

　　ψ（４９）＝４９×（１－１/７）＝４２。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<string>
using namespace std;
int fun(int n){
	int ans=n;
	for(int i=2;i<=sqrt(n);i++){
		if(n%i==0){
			ans=ans/i*(i-1);
			while(n%i==0){
				n=n/i;
			}
		}
	}
	if(n>1){
		ans=ans/n*(n-1);
	}
	return ans;
}
int main(){
	int n;
	while(cin>>n&&n){
		 int ans=fun(n);
		 cout<<ans<<endl;
	}
}