poj2478 Farey Sequence

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

1359

从小于n的数中，选两个互质的数凑成一对，问你能选多少对。有顺序

那么其实就是小于n的互质的数+小于n-1的互质的数+。。。。。其实就是求一个欧拉函数的前缀和。

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <iostream>using namespace std;typedef long long LL;const int MAXN = 1e6+7;bool vis[MAXN];int prim[MAXN];LL phi[MAXN];int tot;void get_phi(){    tot = 0;    for(int i = 2; i < MAXN; ++i)    {        if(!vis[i])        {            prim[tot++] = i;            phi[i] = i-1;        }        for(int j = 0; j < tot; ++j)        {            if(prim[j]*i >= MAXN)break;            vis[i*prim[j]] = 1;            if(i%prim[j] == 0)            {                phi[i*prim[j]] = phi[i]*prim[j];                break;            }            else phi[i*prim[j]] = phi[i]*phi[prim[j]];        }    }    for(int i = 3; i <= 1000000; ++i)phi[i] += phi[i-1];}int main(){    get_phi();    int n;    while(~scanf("%d",&n)&&n)    {        printf("%lld\n",phi[n]);    }    return 0;}