| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 12802 | Accepted: 4998 |
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
/**
题意：求F[i] 的元素的个数，并且0<a<b<=i, gcd(a,b)=1
做法：欧拉 用phi[n]存储小于等于n的素数的个数，然后f[i] = phi[i] + f[i]
**/
#include <iostream>
#include <string.h>
#include <cmath>
#include <algorithm>
#include <stdio.h>
#define maxn 1000100
using namespace std;
long long mmap[maxn];
int phi[maxn];
long long n;
void geteuler()
{
memset(phi,,sizeof(phi));
phi[] = ;
for(int i=; i<=maxn; i++)
{
if(!phi[i])
{
for(int j=i; j<=maxn; j+=i)
{
if(!phi[j])
phi[j] = j;
phi[j] = phi[j]/i*(i-);
}
}
}
}
int main()
{
geteuler();
mmap[] = ;
for(int i=; i<=maxn; i++)
{
mmap[i] = mmap[i-] + phi[i];
}
//freopen("in.txt","r",stdin);
while(~scanf("%lld",&n))
{
if(n == ) break;
printf("%lld\n",mmap[n]);
}
return ;
}