Farey Sequence
| Time Limit: 1000MS |
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Memory Limit: 65536K |
| Total Submissions: 11996 |
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Accepted: 4657 |
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 <= 10
6). 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
欧拉函数：
#include
#include
#define N 1000001 int prime[N]; __int64 ans[N]; int main() { int i,j,n; prime[0]=0; prime[1]=0; for(i=2;i<=N;i++) // 生成素数表 prime[i]=1; for (i=2;i*i<=N;i++) { if(prime[i]) { for(j=i*i;j<=N;j+=i) prime[j]=0; } } for(i=1;i<=N;i++) // 生成欧拉函数值 ans[i]=i; for(i=2;i<=N;i++) { if(prime[i]) { for(j=i;j<=N;j+=i) ans[j]=ans[j]/i*(i-1); // } } for(i=3;i<=N;i++) //前n个数中两两互质的组合数 ans[i]+=ans[i-1]; while(~scanf("%d",&n)&&n!=0) printf("%I64d\n",ans[n]); return 0; }
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