POJ 2478Farey 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

23450

Sample Output

135
9
线性筛求欧拉函数。
#include<set>#include<map>#include<ctime>#include<cmath>#include<stack>#include<queue>#include<bitset>#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<algorithm>#include<functional>#define rep(i,j,k) for (int i = j; i <= k; i++)#define per(i,j,k) for (int i = j; i >= k; i--)#define loop(i,j,k) for (int i = j;i != -1; i = k[i])#define lson x << 1, l, mid#define rson x << 1 | 1, mid + 1, r#define ff first#define ss second#define mp(i,j) make_pair(i,j)#define pb push_back#define pii pair<int,LL>#define in(x) scanf("%d", &x);using namespace std;typedef long long LL;const int low(int x) { return x&-x; }const double eps = 1e-4;const int INF = 0x7FFFFFFF;const int mod = 998244353;const int N = 1e6 + 10;int n;LL sum[N];int phi[N], f[N], p[N], t;void init(){t = 0;sum[1] = phi[1] = f[1] = 1;rep(i, 2, N - 1){if (!f[i]) p[t++] = i, phi[i] = i - 1;for (int j = 0,k; j < t&&p[j] * i < N; j++){k = p[j] * i; f[k] = 1;phi[k] = phi[i] * (p[j] - 1);if (i % p[j] == 0) { phi[k] = phi[i] * p[j]; break; }}sum[i] = sum[i - 1] + phi[i];}}int main(){init();while (scanf("%d", &n), n) printf("%lld\n", sum[n] - 1);return 0;}