递推求欧拉函数

for (i = 1; i <= maxn; i++)
        phi[i] = i;
    for (i = 2; i <= maxn; i += 2)
        phi[i] /= 2;
    for (i = 3; i <= maxn; i += 2)
        if (phi[i] == i)
        {
            for (j = i; j <= maxn; j += i)
                phi[j] = phi[j] / i * (i - 1);
        }

 

例子：poj2478

 

Farey Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11432		Accepted: 4438

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

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;

#define maxn 1000006

long long phi[maxn], num[maxn];
int n;

int main()
{
    //freopen("t.txt", "r", stdin);
    int i, j;
    for (i = 1; i <= maxn; i++)
        phi[i] = i;
    for (i = 2; i <= maxn; i += 2)
        phi[i] /= 2;
    for (i = 3; i <= maxn; i += 2)
        if (phi[i] == i)
        {
            for (j = i; j <= maxn; j += i)
                phi[j] = phi[j] / i * (i - 1);
        }
    num[0] = 0;
    for (int i = 1; i < maxn; i++)
        num[i] = num[i - 1] + phi[i];
    while (scanf("%d", &n), n != 0)
    {
        printf("%lld\n", num[n] - 1);
        //for(i=1;i<5;i++)
        //    cout<<phi[i]<<endl;
    }
    system("pause");
    return 0;
}