[POJ 2407]Relatives(歐拉函數)

[POJ 2407]Relatives(歐拉函數)

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

剛開始我傻X了，本著蒟蒻的思維，照著歐拉函數的公式寫了如下的代碼，先篩素數然後分解質因數最後再算公式

//歐拉函數h(n)=n*(1-1/p1)*(1-1/p2)*'''''*(1-1/pk);
#include 
#include 
#include 
#include 

#define MAXN 1010

using namespace std;

bool isPrime[MAXN];
int Prime[MAXN],PriNum[MAXN],cnt=0,tot=0;

void GetPrime()
{
    cnt=0;
    for(int i=1;i>N;
        if(!N) break;
        DecQualityFactor(N);
        cout<好吧這個代碼根本沒辦法過，因為題目給的n的范圍太大了，數組完爆內存，也就是說這題根本就不能用數組保存什麼質因數和素數的。

下面才是真正的AC代碼，又短又快神神哒


//歐拉函數h(n)=n*(1-1/p1)*(1-1/p2)*'''''*(1-1/pk);
#include 
#include 
#include 
#include 

#define MAXN 1000

using namespace std;

int h(int n)
{
    int ans=n;
    for(int i=2;i<=n;i++)
    {
        if(n%i==0)
        {
            ans=ans/i*(i-1);
            n/=i;
            while(n%i==0) n/=i;
        }
    }
    return ans;
}
int main()
{
    while(1)
    {
        int N;
        cin>>N;
        if(!N) break;
        cout<