欧拉函数 POJ 2407 Relatives&&POJ 2478 Farey Sequence

ZQUOJ 22354&&&POJ 2407  Relatives

Description

Given n, a positive integer, how many positive integers less than n are relatively prime ton ? Two integers a and b are relatively prime if there are no integers x > 1, y > 0, z > 0such 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

有关欧拉函数的知识请参考：http://blog.csdn.net/hillgong/article/details/4214327
题目分析：没什么好讲的，纯粹的欧拉函数模板。

AC代码：

#include<stdio.h>
int main()
{
    int i,n,rea;
    while(scanf("%d",&n)&&n)
    {
        rea=n;
        for(i=2;i*i<=n;i++)
            if(n%i==0)
            {
                rea=rea-rea/i;
                do{
                    n/=i;
                }while(n%i==0);
            }
        if(n>1)
            rea=rea-rea/n;
        printf("%d\n",rea);
    }
    return 0;
}

POJ  2478  Farey Sequence

Description

The Farey Sequence Fn for any integer n with n >= 2 is the set of irreducible rational numbers a/b with 0 < a < b <= n and gcd(a,b) = 1 arranged in increasing order. The first few are 
F2 = {1/2} 
F3 = {1/3, 1/2, 2/3} 
F4 = {1/4, 1/3, 1/2, 2/3, 3/4} 
F5 = {1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5} 

You task is to calculate the number of terms in the Farey sequence Fn.

Input

There are several test cases. Each test case has only one line, which contains a positive integer n (2 <= n <= 10⁶). There are no blank lines between cases. A line with a single 0 terminates the input.

Output

For each test case, you should output one line, which contains N(n) ---- the number of terms in the Farey sequence Fn. 

Sample Input

2
3
4
5
0

Sample Output

1
3
5
9

题意：求2到n的连续欧拉函数的值的和。
分析：因为要频繁地用欧拉函数的值，所以需要预先打表。下面介绍递推求欧拉函数的方法。
可预先置所有数的欧拉函数值为它本身，若p是一个正整数且满足f(p)=p-1，那么p是素数，在遍历过程中如果遇到欧拉函数与自身相等的情况，那么说明该数是素数，把这个数的欧拉函数值改变，同时也把能被该素因子整除的数改变。时间复杂度度为O(nln n)。

AC代码：

#include<stdio.h>
#define MAXN 1000000
double phi[MAXN+10];
int main()
{
    int i,j,n;
    for(i=1;i<=MAXN;i++)    //递推求欧拉函数
        phi[i]=i;
    for(i=2;i<=MAXN;i+=2)
        phi[i]/=2;
    for(i=3;i<=MAXN;i+=2)
        if(phi[i]==i)
        {
            for(j=i;j<=MAXN;j+=i)
                phi[j]=phi[j]/i*(i-1);
        }
    for(i=3;i<=MAXN;i++)  //打表处理FN（2到n的欧拉函数值的和）
        phi[i]+=phi[i-1];
    while(scanf("%d",&n)&&n)
        printf("%.f\n",phi[n]);
    return 0;
}