Farey Sequence
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 14291 | Accepted: 5647 |
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}
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
Source
POJ Contest,Author:Mathematica@ZSU
题解：
直接欧拉函数，求个前缀和，就好了。
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
#define LL long long
LL qz[];
int phi[],prime[],tot;
bool vis[];
void Eular()
{
int i,j;
phi[]=;tot=;
for(i=;i<=;i++)
{
if(vis[i]==false)
{
prime[++tot]=i;
phi[i]=i-;
}
for(j=;j<=tot&&prime[j]*i<=;j++)
{
vis[prime[j]*i]=true;
if(i%prime[j]==)
{
phi[prime[j]*i]=phi[i]*prime[j];
break;
}
phi[prime[j]*i]=phi[prime[j]]*phi[i];
}
}
}
void Qz()
{
for(int i=;i<=;i++)qz[i]=qz[i-]+phi[i];
}
int main()
{
int n;
Eular();
//for(int i=1;i<=100;i++)printf("%d ",phi[i]);
//printf("\n");
Qz();
while()
{
scanf("%d",&n);
if(n==)break;
printf("%lld\n",qz[n]-);
}
return ;
}