POJ 2407 Relatives

Relatives

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11340		Accepted: 5522

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

题目 大意：
       给你一个数n，让你求在n中，和n互质（只有1一个公约数）的数有多少个？

思路 ：
        欧拉公式。
       根据欧拉公式。φ(x)=x(1-1/p1)(1-1/p2)(1-1/p3)(1-1/p4)…..(1-1/pn),

        其中p1, p2……pn为x的所有质因数，x是不为0的整数。

        φ(1)=1（唯一和1互质的数就是1本身）。

        (注意：每种质因数只一个。比如12=2*2*3

         欧拉公式那么φ（12）=12*（1-1/2）*(1-1/3)=4)；

AC 代码：

#include<stdio.h>
int main()
{
int n,sum,i;
double t;
while(~scanf("%d",&n))
{
if(n==0)
{
break;
}
sum=n;
for(i=2;i*i<=n;i++)
{
if(n%i==0)
{
t=1.0/i;
sum=sum*(1-t);
}
while(n%i==0)
{
n=n/i;
}
}
if(n!=1)
{
t=1.0/n;
sum=sum*(1-t);
}
printf("%d\n",sum);
}
return 0;
}