Farey Sequence
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 18635 | Accepted: 7495 |
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
Source
POJ Contest,Author:Mathematica@ZSU
问题简述:(略)
问题分析:
这是一个欧拉函数问题,占个位置不解释。
打表是需要的。计算区间的欧拉函数值,可以依据筛选法的原理进行区间打表。
如果使用欧拉函数,逐个计算i的欧拉函数值phi(i)进行打表,必定导致超时。这里的区间打表,速度要快许多。
程序说明:(略)
题记:(略)
参考链接:(略)
AC的C++语言程序如下:
/* POJ2478 Farey Sequence */
#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
const int N = 1e6;
long long phisum[N + 1];
/* 欧拉函数(筛选法) */
void setphisum()
{
memset(phisum, 0, sizeof(phisum));
for ( int i = 2 ; i <= N ; i ++ )
if ( ! phisum [i] ) {
for ( int j = i ; j <= N ; j += i ) {
if ( ! phisum [j] )
phisum [j ] = j ;
phisum [j] = phisum [j] / i * ( i - 1 ) ;
}
}
for(int i = 2; i <= N; i++)
phisum[i] += phisum[i - 1];
}
int main()
{
setphisum();
int n;
while(~scanf("%d", &n) && n)
printf("%lld\n", phisum[n]);
return 0;
}