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：

 1 #include<iostream>
 2 #include<algorithm>
 3 using namespace std;
 4
 5 int ans;
 6
 7 void solve(int x)
 8 {
 9     ans = x;
10     int ant = 0;
11     for(int i = 2; i*i < x; i++)
12     {
13         if(x%i == 0)
14         {
15             ans = ans / i * (i - 1);
16         }
17         while(x%i == 0) x /= i;
18     }
19
20     if(x > 1)
21         ans = ans/x*(x-1);
22 }
23
24 int main()
25 {
26     int n;
27     while(1)
28     {
29         cin >> n;
30         if(n == 0)
31             break;
32         solve(n);
33         cout << ans << endl;
34     }
35
36
37     return 0;
38 }