POJ 2478 Farey Sequence(Farey数列&欧拉函数求和)

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

思路：

1. 由于Farey数列

包含了

的全部项和与n互质的每个数的相应分数，故有

，

从而

。又由于

，所以要用long long处理。

2. 如何递推地计算φ(n)？——从素数入手

若p|n，那么

若p不整除(n/p)，φ(n)=φ(n/p)*φ(p)=φ(n/p)*(p-1)
若p|(n/p)，则由φ(p^k)=p*φ(p^(k-1))及素因子分解式，得

φ(n)=φ(n/p)*p

*注：此法制表速度比一般的快不少。

完整代码，已优化：

/*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])///若p不整除(n/p)，φ(n)=φ(n/p)*(p-1)
			{
				phi[prime[j] * i] = phi[i] * (prime[j] - 1);
				///不break是因为与p互素的i后面不可能出现(因为i越来越大)，所以要继续算
			}
			else///若p|(n/p)，φ(n)=φ(n/p)*p
			{
				phi[prime[j] * i] = phi[i] * prime[j];
				break;///后面遇到比p1大的p2就不用算了，因为k*p1 * p2 = k*p2 * p1，后面i=k*p2时会算出来的
			}
		}
	}
}

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\n", sum[n]);
	return 0;
}