数学训练----数论F - 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 <= 10 6). 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

本题思路很明显，就是从2开始循环，计算小于等于它们的质数个数之和（求取小于等于一个数的质数个数可以用欧拉函数）。

即：

long long ans=0;for(i=2;i<=n;i++)ans+=oula[i];

但是由于数太大，需要用打表法，不然会超时，因为这个错了好多次。

代码：

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;int oula[1000005];int main(){    int n;    int i,j;    memset(oula,0,sizeof(oula));    for(i=2;i<=1000000;i++)    {        if(!oula[i])        {            for(j=i;j<=1000000;j+=i)            {                if(!oula[j])                    oula[j]=j;                oula[j]=oula[j]/i*(i-1);            }        }    }    while(cin>>n,n)    {        long long ans=0;        for(i=2;i<=n;i++)           ans+=oula[i];        cout<<ans<<endl;    }    return 0;}