POJ 2478-Farey Sequence (Euler's function of Evaluate Method)

POJ 2478-Farey Sequence (Euler's function of Evaluate Method)

 

Farey Sequence Time Limit:1000 MS Memory Limit:65536KB 64bit IO Format:% I64d & % I64u Submit Status Practice POJ 2478 Appoint description: System Crawler)

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 shoshould output one line, which contains N (n) ---- the number of terms in the Farey sequence Fn.

Sample Input

23450

Sample Output

1359

 

Given a number of n, calculate the number of real scores composed of two or more numbers that are less than or equal to n.

Idea: This blog has reasoning about this question ----> click to open the link

 

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          using namespace std;typedef long long LL;const int inf=0x3f3f3f3f;const double pi= acos(-1.0);LL phi[1000010];LL res[1000010];void Euler(){ int i,j; memset(phi,0,sizeof(phi)); phi[1]=1; for(i=2;i<=1000010;i++) { if(!phi[i]) { for(j=i;j<=1000010;j+=i) { if(!phi[j]) phi[j]=j; phi[j]=phi[j]/i*(i-1); } } }}int main(){ int n; Euler(); memset(res,0,sizeof(res)); res[1]=res[2]=1; for(int i=3;i<1000010;i++) res[i]=res[i-1]+phi[i]; while(~scanf("%d",&n)){ if(!n) break; printf("%lld\n",res[n]); } return 0;}