Relatives（容斥）

Relatives

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15708		Accepted: 7966

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

题解：通过这道题了解到了欧拉函数

欧拉函数可以求出小于n的质因数，则这题可以通过公式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）。来求解。例如：

12=2*2*3 

那么φ（12）=12*（1-1/2）*(1-1/3)=4)

AC代码：

#include<iostream>
#include<algorithm>
using namespace std;

int ans;

void solve(int x)
{
    ans = x;
    int ant = 0;
    for(int i = 2; i*i < x; i++)
    {
        if(x%i == 0)
        {
            ans = ans / i * (i - 1);
        }
        while(x%i == 0) x /= i;
    }

    if(x > 1)
        ans = ans/x*(x-1);
}

int main()
{
    int n;
    while(1)
    {
        cin >> n;
        if(n == 0)
            break;
        solve(n);
        cout << ans << endl;
    }


    return 0;
}