POJ 2478 Farey Sequence 欧拉函数运用

Farey Sequence
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu
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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

#include<stdio.h>#include<string>#include<cstring>#include<queue>#include<algorithm>#include<functional>#include<vector>#include<iomanip>#include<math.h>#include<iostream>#include<sstream>#include<stack>#include<set>#include<bitset>using namespace std;const int INF=0x3f3f3f3f;const int MAX=1000005;int Phi[MAX],N;void Init(){    memset(Phi,0,sizeof(Phi));    for (int i=2; i<MAX; i++)        if (!Phi[i])            for (int j=i; j<MAX; j+=i)            {                if (!Phi[j])                    Phi[j]=j;                Phi[j]=Phi[j]/i*(i-1);            }}int main(){    cin.sync_with_stdio(false);    Init();    while (cin>>N&&N)    {        long long Ans=0;        for (int i=2;i<=N;i++)            Ans+=Phi[i];        cout<<Ans<<endl;    }    return 0;}