Time Limit : 2000/1000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)
Total Submission(s) : 4 Accepted Submission(s) : 4
Problem 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
求小于等于n且与n互质的数
定理：
设A, B, C是跟m, n, mn互质的数的集，据中国
剩余定理，A*B和C可建立一一对应的关系。因此φ(n)的值使用
算术基本定理便知，
若
则
例如
与欧拉定理、
费马小定理的关系
对任何两个互质的正整数a, m(m>=2)有
即欧拉定理
当m是质数p时，此式则为：
即费马小定理。
代码按照定理来写的（不能用线性筛因为复杂度1e9
#include <iostream>
#include <string>
#include <string.h>
#include <algorithm>
#include <cmath>
#include <vector>
#include <stdio.h>
#include <stack>
using namespace std;
int eular(int num) {
int ans = 1;
for (int i = 2; i * i <= num; ++i) {
if (num % i == 0) {
num /= i;
ans *= i - 1;
while (num % i == 0) {
num /= i;
ans *= i;
}
}
}
if (num > 1)
ans *= (num - 1);
return ans;
}
int main() {
//freopen("1.txt", "r", stdin);
int n;
while (cin >> n && n) {
cout << eular(n) << endl;
}
}
