Farey Sequence

Time Limit: 1000MS		Memory Limit: 65536K

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

23450

Sample Output

1359

Source

,Author:Mathematica@ZSU

 

欧拉函数模板题，16ms的O(n)线性筛，代码是抄贾教的。。

Codes:

1 #include
      
        2 #include
       
         3 using namespace std; 4 int n,phi[1001000],tot,prime[500000]; 5 long long sum[1001000]; 6 bool check[1000100]; 7 void PHI(int n){ 8     phi[1] = 1; 9     for(int i=2;i<=n;i++){10         if(!check[i]){11             prime[++tot] = i;12             phi[i] = i - 1;13         }14         for(int j=1;j<=tot;j++){15             if(prime[j]*i>n) break;16             check[prime[j]*i] = true;17             if(i%prime[j]==0){18                 phi[i*prime[j]] = phi[i] * prime[j];19                 break;20             }else   phi[i*prime[j]] = phi[i] * (prime[j]-1);21         }22     }23 }24 25 int main(){26     PHI(1000000);27     sum[2] = 1;28     for(int i=3;i<=1000000;i++) sum[i]+=sum[i-1] + phi[i];29     while(scanf("%d",&n)!=EOF && n) cout<
        
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