Relatives 和 GCD Again 【欧拉函数】

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
2
4
0
Sample Output
0
1
代码

#include<bits/stdc++.h>using namespace std ;typedef long long LL ;const int MAXN = 1e3 +10;const int MAXM = 1e5 ;const int mod  = 1e9+7 ;const double pi=acos(-1.0);int eular(int n){    int ans=n;    for(int i=2;i*i<=n;i++){        if(n%i==0){            ans=ans/i*(i-1);            while(n%i==0) n/=i;        }    }    if(n>1) ans=ans/n*(n-1);  // 如果大于1那么一定是个素数    return ans;}int main(){     int n;    while(~scanf("%d",&n)&&n){        printf("%d\n",n-1-eular(n));    }     return  0;}

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
代码

#include<cstdio>using namespace std ;typedef long long LL ;const int MAXN = 1e3 +10;const int MAXM = 1e5 ;const int mod  = 1e9+7 ;int eular(int n){    int ans=n;    for(int i=2;i*i<=n;i++){        if(n%i==0) {            ans=ans/i*(i-1);            while(n%i==0) n/=i;        }    }    if(n>1) ans=ans/n*(n-1);    return ans;}int main(){     int n;    while(scanf("%d",&n)&&n){        printf("%d\n",eular(n));    }    return  0;}