76-Relatives-欧拉函数

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

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　Relatives

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16652		Accepted: 8470

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

 

思路：欧拉函数板子题，就是就给定数字的欧拉函数

 

欧拉函数详解

 

#include <iostream>
using namespace std;
#define LL long long
LL Phi(LL x){   //按定义直接算
	if(x == 1){
		return 1;
	}
	LL ans = x;
	for(int i = 2; i * i <= x; i++){
		if(x % i == 0){
			ans = ans / i * (i - 1);
			while(x % i == 0){
				x = x / i;
			}
		}
	}
//	cout << ans << " " << x << endl;
	if(x >= 2)
		ans = ans / x * (x - 1);
	return ans;
}

//bool visit[10001];
//int tot=0, pri[1000005], phi[1000005];
//LL Phi(LL n){   //线性筛选
//	phi[1] = 1;
//	for(int i = 2; i <= n; i++){
//		if(!visit[i]){
//			pri[++tot] = i;
//			phi[i] = i - 1;
//		}
//		for(int j = 1, x; j <= tot && (x = i * pri[j]) <= n; j++){
//			visit[x] = true;
//			if(i % pri[j] == 0){  //不互质时
//				phi[x] = phi[i] * pri[j];
//				break;
//			}
//			else{
//				phi[x] = phi[i] * phi[pri[j]];
//			}
//		}
//	}
//	return phi[n];
//} 

int main(){
	LL x;
	while(cin >> x && x){
		cout << Phi(x) << endl;
	}
	return 0;
}