题目链接：http://poj.org/problem?id=2478
Farey Sequence
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 19736 | Accepted: 7962 |
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}
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
题目大意：很容易可以发现是求2-n的所有数的欧拉函数值之和
看代码：
/**
有三条特性
若a为质数 phi[a]=a-1
若a为质数,b%a==0 phi[a*b]=phi[b]*a;
若a b 互质 phi[a*b]=phi[a]*phi[b]（当a为质数 如果b%a!=0） */
#include<iostream>
#include<cstdio>
using namespace std;
typedef long long LL;
const int maxn=1e6+;
int phi[maxn],prime[maxn],p[maxn];//phi[i]代表i的欧拉函数值 prime[i]=0代表是素数 1代表不是素数 p存储素数
void make()
{
phi[]=;//特例
int num=;
for(int i=;i<=maxn;i++)
{
if(!prime[i])//是素数
{
p[num++]=i;//
phi[i]=i-;//素数的欧拉函数值就是它的值减1
}
for(int j=;j<num&&p[j]*i<maxn;j++)//用当前已经得到的素数筛去p[j]*i
{
prime[p[j]*i]=;//可以确定p[j]*i不是质数
if(i%p[j]==)//第二条特性
{
phi[p[j]*i]=phi[i]*p[j];
break;//欧拉筛的核心语句 保证每个数只会被自己最小的质因子筛掉一次
}
else phi[p[j]*i]=phi[i]*phi[p[j]];
}
}
return ;
} int main()
{
make();
int n;
while(scanf("%d",&n)!=EOF)
{
if(n==) break;
LL sum=;
for(int i=;i<=n;i++) sum+=phi[i];
printf("%lld\n",sum);
}
// for(int i=1;i<=100;i++) cout<<phi[i]<<" "; return ;
}