GCD Again(欧拉函数）

GCD Again

Time Limit : 1000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 13 Accepted Submission(s) : 6

Problem Description

Do you have spent some time to think and try to solve those unsolved problem after one ACM contest?
No? Oh, you must do this when you want to become a "Big Cattle".
Now you will find that this problem is so familiar:
The greatest common divisor GCD (a, b) of two positive integers a and b, sometimes written (a, b), is the largest divisor common to a and b. For example, (1, 2) =1, (12, 18) =6. (a, b) can be easily found by the Euclidean algorithm. Now I am considering a little more difficult problem:
Given an integer N, please count the number of the integers M (0<M<N) which satisfies (N,M)>1.
This is a simple version of problem “GCD” which you have done in a contest recently,so I name this problem “GCD Again”.If you cannot solve it still,please take a good think about your method of study.
Good Luck!

Input

Input contains multiple test cases. Each test case contains an integers N (1<N<100000000). A test case containing 0 terminates the input and this test case is not to be processed.

Output

For each integers N you should output the number of integers M in one line, and with one line of output for each line in input.

Sample Input

240

Sample Output

01

Author

lcy

Source

2007省赛集训队练习赛（10）_以此感谢DOOMIII

题意：
          求小于n的gcd(i,n)大于1的个数,即求除n外与n有共同约数i的个数(0<i<n)

解题思路：欧拉函数

欧拉函数的介绍：
在数论，对正整数n，欧拉函数是少于或等于n的数中与n互质的数的数目。此函数以其首名研究者欧拉命名，它又称为Euler's totient function、φ函数、欧拉商数等。 例如φ(8)=4，因为1,3,5,7均和8互质。 从欧拉函数引伸出来在环论方面的事实和拉格朗日定理构成了欧拉定理的证明。
φ函数的值通式：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

　　那么φ（12）=12*（1-1/2）*(1-1/3)=4)

　　若n是质数p的k次幂，φ(n)=p^k-p^(k-1)=(p-1)p^(k-1)，因为除了p的倍数外，其他数都跟n互质。

　　欧拉函数是积性函数——若m,n互质，φ(mn)=φ(m)φ(n)。

　　特殊性质：当n为奇数时，φ(2n)=φ(n), 证明于上述类似。

#include<cstdio>using namespace std;int f(int x){    int i,t=1;    for(i=2;i*i<=x;i++)    {        if(x%i==0)        {            x/=i;            t*=i-1;            while(x%i==0)            {                x/=i;                t*=i;            }        }    }    if(x>1)     t*=x-1;    return t;}int main(){    int n;    while(scanf("%d",&n),n)     printf("%d\n",n-1-f(n));    return 0;}