Farey Sequence
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 17894 | Accepted: 7179 |
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.
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
分析：这题很明显的欧拉函数啊，F(N)就是欧拉函数的前N项和(不包括1)。
#include<cstdio> long long a[1000100]; void Euler() { for(int i=1;i<1000001;i++) a[i]=i; for(int i=2;i<1000001;i++) { if(a[i]==i) { for(int j=i;j<1000001;j+=i) a[j]=a[j]/i*(i-1); } } for(int i=3;i<1000001;i++) a[i]+=a[i-1];//从3开始 } int main() { int N; Euler(); while(scanf("%d",&N)&&N) printf("%lld\n",a[N]); return 0; }
