POJ 2407-Relatives (欧拉函数)

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

7120

 

Sample Output

64

 

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;    }}