poj2407解题报告

Relatives

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6232		Accepted: 2786

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

题目大意：给定一个数(小于1,000,000,000),求与他互质数的个数。。

思路：利用欧拉函数，将n素因子分解，带入欧拉函数求值。。

#include<iostream>
#include<cmath>
using namespace std;
__int64 o(__int64 a,__int64 b)
{
	if(!a)
		return b;
	return o(b%a,a);
}
int main()
{
	__int64 i,n,o1,o2,gys;
	while(scanf("%I64d",&n)&&n)
	{
		if(n==1)
		{	cout<<'0'<<endl;	continue;	}
		o1=n;
		o2=1;
		for(i=2;i<=(int)sqrt((double)n);i++)
			if(n%i==0)
			{
				o1*=(i-1);
				o2*=i;
				gys=o(o1,o2);
				o1/=gys;	o2/=gys;
				while(n%i==0)
					n/=i;
			}
		if(n!=1)
		{
			o1*=(n-1);
			o2*=n;
			gys=o(o1,o2);
			o1/=gys;	o2/=gys;
		}
			printf("%I64d/n",o1);
	}

	return 0;
}