| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 16494 | Accepted: 8368 |
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
利用歐拉函數和它本身不同質因數的關係，用篩法計算出某個範圍內所有數的歐拉函數值。
歐拉函數和它本身不同質因數的關係：歐拉函數ψ（Ｎ）＝Ｎ｛∏ｐ｜Ｎ｝（１－１/ｐ）。（Ｐ是數Ｎ的質因數）
如：
ψ（１０）＝１０×（１－１/２）×（１－１/５）＝４；
ψ（３０）＝３０×（１－１/２）×（１－１/３）×（１－１/５）＝８；
ψ（４９）＝４９×（１－１/７）＝４２。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<string>
using namespace std;
int fun(int n){
int ans=n;
for(int i=2;i<=sqrt(n);i++){
if(n%i==0){
ans=ans/i*(i-1);
while(n%i==0){
n=n/i;
}
}
}
if(n>1){
ans=ans/n*(n-1);
}
return ans;
}
int main(){
int n;
while(cin>>n&&n){
int ans=fun(n);
cout<<ans<<endl;
}
}