POJ2487 Farey Sequence 【欧拉函数线性筛选模板题】

Farey Sequence Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 18838   Accepted: 7581 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

题意：求每个小于等于n的数的欧拉函数相加的和

思路：先求每个数的欧拉函数然后相加，注意此题从i=2开始，若从一开始结果减一

代码：

#include
#include
using namespace std;
const int maxn=1000008;
int phi[maxn],prime[maxn],tot;
int mark[maxn];
void getph()//欧拉函数的线性筛选打表
{
    phi[1]=1;
    for(int i=2;i<=maxn;i++)
    {
        if(!mark[i])
        {
            prime[++tot]=i;
            phi[i]=i-1;
        }
        for(int j=1;j<=tot;j++)
        {
            if(i*prime[j]>maxn) break;
            mark[i*prime[j]]=1;
            if(i%prime[j]==0)
            {
                phi[i*prime[j]]=prime[j]*phi[i];
                break;
            }
            else
            {
                phi[i*prime[j]]=phi[i]*(prime[j]-1);
            }
        }
    }
}
int main()
{
    long long n,ans;
    getph();
    while(scanf("%lld",&n)!=EOF)
    {
        if(n==0)
          break;
          ans=0;
        for(int i=2;i<=n;i++)
        {
               ans+=phi[i];
        }
        printf("%lld\n",ans);
    }
    return 0;
}