POJ 2407 素数筛

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
素数筛水题，本题我用的是一种据说挺厉害的素数筛，特写个博客记录下来，下附ac代码：

#include<iostream>

#include<cmath>

#include<cstdio>

#include<cstring>

using namespace std;

int pri[400100];

int p[400000];

void prime()

{

    memset(pri,0,sizeof(pri));

    pri[0]=pri[1]=1;

    int num=0,i,j;

    for(i = 2; i < 400100; ++i)

    {

        if(!(pri[i])) p[num++] = i;

        for(j = 0; (j<num && i*p[j]<400100); ++j)

        {

            pri[i*p[j]] = 1;

            if(!(i%p[j])) break;

        }

    }

}

int power(int x,int y)

{

    int mi=1;

    for(int i=1;i<=y;i++)

    {

        mi*=x;

    }

    return mi-mi/x;

}

int main()

{

    int n;

    prime();

    while(scanf("%d",&n),n)

    {

        int temp=n,sum=1,ok=0;

        for(int i=0;i<400000;i++)

        {

            if(temp%p[i]==0)

            {

                int tot=0;

                while(temp%p[i]==0)

                {

                    tot++;

                    temp/=p[i];

                }

                sum*=power(p[i],tot);

            }

            else if(sum==0&&p[i]>=(int)(sqrt(n)+1))

            {

                printf("%d\n",n-1);

                break;

            }

            else if(temp==1)

            {

                ok=1;break;

            }

        }

        if(ok)printf("%d\n",sum);

    }

    return 0;

}