POJ 2478 Farey Sequence

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

23450

Sample Output

1359

Source

POJ Contest,Author:Mathematica@ZSU

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欧拉函数，相当于是求2~n的欧拉函数之和~

今天被long long坑了两次……

#include<cstdio>#define ll long longint n;ll a[1000001],phi[1000001];bool b[1000001];void findd(ll u){for(ll i=2;i<=u;i++){if(!b[i]){a[++a[0]]=i;phi[i]=i-1;}for(ll j=1;a[j]*i<=u;j++){b[a[j]*i]=1;if(i%a[j]==0){phi[a[j]*i]=phi[i]*a[j];break;}phi[a[j]*i]=phi[i]*(a[j]-1);}}}int main(){findd(1000000);for(int i=3;i<=1000000;i++) phi[i]+=phi[i-1];while(scanf("%d",&n)==1 && n) printf("%lld\n",phi[n]);return 0;}