poj 2478 Farey Sequence 欧拉函数递推打表

Farey Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14434		Accepted: 5719

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 <= 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

题意：求f(n)中数的个数。
思路：可以看出f(n)=f(n-1)+phi(n)（法雷数列中分母为n的项，其分子与其互质，所以分母为n的数的个数即为phi(n)），打表求出1e6内的欧拉函数，累加即可

代码：
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <map>
using namespace std;
long long p[1000006];
void phi(long long N)
{
for(long long i=1; i<N; i++) p[i] = i;
for(long long i=2; i<N; i+=2) p[i] >>= 1;
for(long long i=3; i<N; i+=2)
{
if(p[i] == i)
{
for(long long j=i; j<N; j+=i)
p[j] = p[j] - p[j] / i;
}
}
}
int main(int argc, char const *argv[])
{ long long i,j,k,m,n,t;
long long sum=0;
phi(1000001);
while(~scanf("%lld",&n))
{ sum=0;
if(n==0)break;
for(i=2;i<=n;i++)sum+=p[i];
printf("%lld\n",sum);

}
return 0;
}