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：

View Code

 1 #include<stdio.h>
 2 int main()
 3 {
 4     int i,n,rea;
 5     while(scanf("%d",&n)&&n)
 6     {
 7         rea=n;
 8         for(i=2;i*i<=n;i++)
 9             if(n%i==0)
10             {
11                 rea=rea-rea/i;
12                 do{
13                     n/=i;
14                 }while(n%i==0);
15             }
16         if(n>1)
17             rea=rea-rea/n;
18         printf("%d\n",rea);
19     }
20     return 0;
21 }

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

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：

View Code

 1 #include<stdio.h>
 2 #define MAXN 1000000
 3 double phi[MAXN+10];
 4 int main()
 5 {
 6     int i,j,n;
 7     for(i=1;i<=MAXN;i++)    //递推求欧拉函数
 8         phi[i]=i;
 9     for(i=2;i<=MAXN;i+=2)
10         phi[i]/=2;
11     for(i=3;i<=MAXN;i+=2)
12         if(phi[i]==i)
13         {
14             for(j=i;j<=MAXN;j+=i)
15                 phi[j]=phi[j]/i*(i-1);
16         }
17     for(i=3;i<=MAXN;i++)  //打表处理FN（2到n的欧拉函数值的和）
18         phi[i]+=phi[i-1];
19     while(scanf("%d",&n)&&n)
20         printf("%.f\n",phi[n]);
21     return 0;
22 }