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;
}
0

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