Relatives（poj2407）（求大数的欧拉函数模板题）

发布时间：2020-12-14 02:19:38 所属栏目：大数据来源：网络整理

导读：Link：http://poj.org/problem?id=2407 Relatives Time Limit: ?1000MS ? Memory Limit: ?65536K Total Submissions: ?12582 ? Accepted: ?6144 Description Given n,a positive integer,how many positive integers less than n are relatively prime to n?

Link：http://poj.org/problem?id=2407

Relatives

Time Limit:?1000MS	?	Memory Limit:?65536K
Total Submissions:?12582	?	Accepted:?6144

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. 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

AC code：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
#include<map>
#include<stack>
#include<vector>
#define LL long long
#define MAXN 100010
using namespace std;
//直接求解欧拉函数
int euler(int n){ //返回euler(n) 
     int res=n,a=n;
     for(int i=2;i*i<=a;i++){
         if(a%i==0){
             res=res/i*(i-1);//先进行除法是为了防止中间数据的溢出 
             while(a%i==0) a/=i;
         }
     }
     if(a>1) res=res/a*(a-1);
     return res;
}

int main()
{
	int n;
	while(~scanf("%d",&n))
	{
		if(n==0)
			break;
		printf("%dn",euler(n));
	 } 
  	return 0;
}

（编辑：李大同）

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