Relatives

Relatives

Time Limit(Common/Java):1000MS/10000MS     Memory Limit:65536KByte
Total Submit: 9            Accepted: 4

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

分析：

用欧拉函数做。对于正整数n，欧拉函数可以求出在小于等于n的数中与n互质的数的个数。

欧拉函数：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

那么φ（12）=12*（1-1/2）*(1-1/3)=4)

 

代码：

#include<stdio.h>
#include<math.h>
int euler(long n)
{
    long res=n, x=n;
    long i,tmp=sqrt(n);
    for (i=2; i<=tmp; i++)
    {
        if (x%i==0)
        {
            res = res/i*(i-1);
        }
        while (x%i==0) x/=i;    //去除x中所有的质因数i,确保下次计算时i是质因数。
    }
    if (x>1) res = res/x*(x-1);    //当n是由两个质因数组成的时候，条件成立
    return res;
}
int main()
{
    long n;
    while (scanf("%ld", &n), n)
    {
        printf("%ld\n", euler(n));
    }
    return 0;
}