POJ 2478 Farey Sequence(Farey数列&Euler関数加算)

Farey Sequence
http://poj.org/problem?id=2478
Time Limit:
 1000MS
Memory Limit: 65536K
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

考え方:
1.Farey数列に含まれる全ての項とnとの相互作用の各数に対応する点数であるため、
,
それで
.
また、そのためlong longで処理します.
2.繰返し計算方法φ(n)?——素数から
p|nなら
pが完全に除去されなければ(n/p)、φ(n)=φ(n/p)*φ(p)=φ(n/p)*(p-1)p|(n/p)の場合、φ(p^k)=p*φ(p^(k-1))及び素因子分解式
φ(n)=φ(n/p)*p
*注:この法制表の速度は一般よりずっと速いです.
完全なコード、最適化:

/*32ms,13204KB*/

#include <cstdio>
const int len = 1000002;
const int le = 78498;

int prime[le], phi[len];
bool unprime[len];
long long sum[len];

inline void Euler()
{
	int i, j, k = 0;
	for (i = 2; i < len; i++)
	{
		if (!unprime[i])
		{
			prime[k++] = i;
			phi[i] = i - 1;
		}
		for (j = 0; j < k && prime[j] * i < len; j++)
		{
			unprime[prime[j] * i] = true;
			if (i % prime[j])/// p   (n/p)，φ(n)=φ(n/p)*(p-1)
			{
				phi[prime[j] * i] = phi[i] * (prime[j] - 1);
				/// break    p   i       (  i    )，      
			}
			else/// p|(n/p)，φ(n)=φ(n/p)*p
			{
				phi[prime[j] * i] = phi[i] * prime[j];
				break;///     p1  p2     ，  k*p1 * p2 = k*p2 * p1，  i=k*p2      
			}
		}
	}
}

int main()
{
	Euler();
	int i, n;
	for (i = 2; i < len; i++)
		sum[i] = sum[i - 1] + phi[i];
	while (scanf("%d", &n), n)
		printf("%lld
", sum[n]);
	return 0;
}