poj 2478 Farey Sequence（欧拉函数）

Farey Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13204		Accepted: 5181

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

Source
POJ Contest,Author:Mathematica@ZSU

简单的欧拉函数模板题。
所谓欧拉函数：对于一个正整数n，小于n且和n 互质的正整数（包括1 ）的个数，记做φ(n) 。
通式：φ(x)=x*(1-1/p1)*(1-1/p2)*(1-1/p3)*(1-1/p4)…..(1-1/pn),其中p1,
p2……pn为x的所有质因数，x是不为0的整数。φ(1)=1（唯一和1互质的数就是1本身）。

欧拉函数代码实现：

//直接求解欧拉函数
int euler(int n){ //返回euler(n)
int res=n,a=n;
for(int i=2;i*i<=a;i++){
if(a%i==0){
res=res/i*(i-1);//先进行除法是为了防止中间数据的溢出
while(a%i==0) a/=i;
}
}
if(a>1) res=res/a*(a-1);
return res;
}

//筛选法打欧拉函数表
#define Max 1000001
int euler[Max];
void Init(){
euler[1]=1;
for(int i=2;i<Max;i++)
euler[i]=i;
for(int i=2;i<Max;i++)
if(euler[i]==i)
for(int j=i;j<Max;j+=i)
euler[j]=euler[j]/i*(i-1);//先进行除法是为了防止中间数据的溢出
}

本题就是欧拉函数的直接使用：

#include<stdio.h>
#include<string.h>
#include<math.h>
#define LL  __int64
#define Max 1005000
LL sum[1005000];
void init(){
sum[1]=1;
for(LL i=2;i<Max;i++)
sum[i]=i;
for(LL i=2;i<Max;i++)
if(sum[i]==i)
for(LL j=i;j<Max;j+=i)
sum[j]=sum[j]/i*(i-1);
}

int main()
{
LL n;
LL i,t;
init();
while(scanf("%I64d",&n)!=EOF)
{
t=0;
if(n==0)break;
for(i=2;i<=n;i++)
t+=sum[i];
printf("%I64d\n",t);
}
return 0;
}